Chapters 15 & 16
Ammonia, NH3, is a weak base with a Kb value of 1.8×10^−5. What is the percent ionization of ammonia at this concentration? Express your answer with the appropriate units. (0.250 M)
% ionization = 0.849% (My answer of 0.84% was accepted.) The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B B(aq)+H2O(l)⇌BH+(aq)+OH−(aq) this constant is given by Kb=[BH+][OH−] / [B] Strong bases will have a higher Kb value. Similarly, strong bases will have a higher percent ionization value. Strong bases ionize 100%, but weak bases only partially ionize. The concentration of hydroxide ions represents the amount of ammonia molecules that ionized. Express that as a percentage of the initial amount of ammonia: % ionization = ([OH-]eq / initial [NH3]) x 100% Find the equilibrium OH- concentration.
At 2000 ∘C the equilibrium constant for the reaction 2NO(g)⇌N2(g)+O2(g) is Kc=2.4×10^3. If the initial concentration of NO is 0.325 M, what is the equilibrium concentration of NO? Express your answer to one significant figure and include the appropriate units.
0.003 M (Don't forget that the concentration of NO needs to be squared because it has a coefficient of 2) Use an equilibrium table to represent the changes in concentration over the course of the reaction. Since NO has two molar equivalences in the balanced equation, it undergoes a change of {−2x}, and the change in N2 and O2 is +x. The equilibrium changes are tabulated below: __________ 2NO(g) _______ N2(g) _______ O2(g) Initial: _____ 0.325 __________ 0 ___________ 0 Change: ____ −2x __________ +x __________ +x Final: _____ 0.325−2x ________ x ___________ x Using the variables, an algebraic expression is written for Kc. Kc = [x][x] / [0.325−2x]^2 = 2.4×10^3 The expression is simplified by combining the x terms in the numerator and taking the square root of both sides. [x] / [0.325−2x] = (2.4×103)^.5 (<-- a power to .5 is the same as sqrt) The x in the numerator is isolated by multiplying both sides by [0.325−2x]. [x] = (2.4×10^3)^.5 * (0.325−2x) The terms on the right side of the equation are distributed. x = 15.92 − 98x The equation is further simplified by adding 98x to each side. 99x = 15.92 Finally, both sides are divided by 99. x = 0.1608 Since NO undergoes a change of −2x, the equilibrium concentration of NO is 0.325−2x, and x is equivalent to 0.1608 M. [NO]equil. = 0.325 − 2(0.1608) = 0.003 M
The equilibrium constant, Kc, for the reaction N2O4 (g) ⇄ 2NO2 (g) is 0.211 at 100°C. If the equilibrium concentration of N2O4 is 0.00251 M, calculate the equilibrium concentration of NO2.
0.0230 M The equilibrium constant expression for the reaction is: Kc = [NO2]^2 / [N2O4].
What is the pH of a solution made by diluting 25 mL of 6.0 M HCl(aq) to a total volume of 0.200 L?
0.12 When the solution is diluted, the [H+] concentration will decrease to 0.75 M, and when this value is plugged into Equation 16.17, we see that the pH = 0.12.
At 2000 ∘C the equilibrium constant for the reaction 2NO(g)⇌N2(g)+O2(g) is Kc=2.4×10^3. If the initial concentration of NO is 0.325 M, what is the equilibrium concentration of N2? Express your answer to two significant figures and include the appropriate units.
0.16 M (See the card showing how to solve for NO if you need help understanding where this number came from. It is the value for x.)
At 2000 ∘C the equilibrium constant for the reaction 2NO(g)⇌N2(g)+O2(g) is Kc=2.4×10^3. If the initial concentration of NO is 0.325 M, what is the equilibrium concentration of O2? Express your answer to two significant figures and include the appropriate units.
0.16 M (See the card showing how to solve for NO if you need help understanding where this number came from. It is the value for x.)
A certain weak acid, HA, has a Ka value of 9.2×10^−7. Calculate the percent ionization of HA in a 0.10 M solution. Express your answer as a percent using two significant figures.
0.30 % Percent ionization for a weak acid (HA) is determined by the following formula: Percent ionization = [HA]ionized / [HA]initial × 100% For strong acids, ionization is nearly complete (100%) at most concentrations. However, for weak acids, the percent ionization changes significantly with concentration. The more diluted the acid is, the greater percent ionization. (Make an ICE table with 0.1 as the initial concentration of HA and use it with Ka to solve for x which is the value for [HA]ionized. Then plug that into the percent ionization equation.)
A certain weak acid, HA, has a Ka value of 9.2×10^−7. Calculate the percent ionization of HA in a 0.010 M solution. Express your answer as a percent using two significant figures.
0.96 % (My answer of 0.95% was accepted.) Comparing the answers to parts A and B, you can see that the percent ionization of a weak acid increases as the concentration decreases. Percent ionization for a weak acid (HA) is determined by the following formula: Percent ionization = [HA]ionized / [HA]initial × 100% For strong acids, ionization is nearly complete (100%) at most concentrations. However, for weak acids, the percent ionization changes significantly with concentration. The more diluted the acid is, the greater percent ionization. (Make an ICE table and use it with Ka to solve for x which is the value for [HA]ionized. Then plug that into the percent ionization equation.)
A 0.15 M solution of imidazole, C3N2H4, a weak organic base, has a pH of 10.11. What is the Kb of imidazole?
1.1 × 10^-7 From the pH you've correctly determined [OH-] = [C2N2H5+] = 1.3 × 10^-4. By using these values in Equation 16.34, you have obtained the correct value of Kb.
Consider the reaction CaSO4(s)⇌Ca2+(aq)+SO4^2−(aq) At 25 ∘C the equilibrium constant is Kc=2.4×10^−5 for this reaction. If the resulting solution has a volume of 1.9 L , what is the minimum mass of CaSO4(s) needed to achieve equilibrium? Express your answer to two significant figures and include the appropriate units.
1.3 g (Use the equilibrium expression to find the eq. concentration of either products. Notice that CaSO4 is a solid and so isn't included. The EC of either product will be the concentration of CaSO4 needed. Multiply the [CaSO4] by 1.9 liters to find how many moles of CaSO4 you need then convert moles to grams.)
At 300°C, Kc for the following reaction is 31. Calculate Kp. PCl5 (g) ⇄ PCl3 (g) + Cl2 (g).
1.5 × 10^3 Kc and Kp are related by the equation: Kp=Kc(RT)∆n. See Section 15.2
What is the pH of a 7.5 × 10-3 M Ba(OH)2 solution at 25°C?
12.18 Ba(OH)2 is a strong base and; therefore, a strong electrolyte. See Section 16.5 (There is 7.5E-3 of the substance, but each mole of the substance dissociates into 2 moles of OH- which means there is a concentration of .015 M of OH-. Do pOH = -log[OH-] then pH + pOH = 14.)
When an unknown substance is dissolved in water at 25 °C, the concentration of OH− ions is determined to be 2.5 × 10^−9 M. From this we can conclude that [H3O+] = ________ and that the solution is ________. 1.0 × 10−7 M, acidic 2.5 × 10−9 M, neutral 4.0 × 10−6 M, acidic 4.0 × 10−6 M, basic
4.0 × 10−6 M, acidic (Find the concentration of H3O+ using the equation [H+][OH-] = 1.0E-14. Then you can either compare that number directly to the concentration of OH- given in the problem to see which one is larger or plug [H+] into pH = -log [H+].)
Consider the reaction CaSO4(s)⇌Ca2+(aq)+SO4^2−(aq) At 25 ∘C the equilibrium constant is Kc=2.4×10^−5 for this reaction. If excess CaSO4(s) is mixed with water at 25 ∘C to produce a saturated solution of CaSO4, what is the equilibrium concentration of Ca2+? Express your answer to two significant figures and include the appropriate units.
4.9×10^−3 M (Use the equilibrium expression. Notice that CaSO4 is a solid and so isn't included.)
Consider the reaction CaSO4(s)⇌Ca2+(aq)+SO4^2−(aq) At 25 ∘C the equilibrium constant is Kc=2.4×10^−5 for this reaction. If excess CaSO4(s) is mixed with water at 25 ∘C to produce a saturated solution of CaSO4, what is the equilibrium concentration of SO4^2−? Express your answer to two significant figures and include the appropriate units.
4.9×10^−3 M (Use the equilibrium expression. Notice that CaSO4 is a solid and so isn't included.)
What is the Ka of an unknown weak acid HA, at 25°C, if the pH of a 2.5 × 10^-2 M solution of the acid was measured and found to be 4.94?
5.3 × 10^-9 (HA --> H+ + A- Use the pH to find [H+]eq. This value will also be the value for [A-]eq as well because every mol of H+ that is dissociated, one mol of the conjugate base is formed as well. Lastly we are given [HA]eq in the problem. Plug the equilibrium concentrations into the equation: Ka = [H+][A-] / [HA])
Determine the concentration of [H+] in a HCl solution with a pH of 3.2?
6.3 x 10^-4 M (pH = -log[H+] there to find [H+] do 10^-[H+])
The Ka for hypochlorous acid, HOCl, is 3.0 × 10^-8 at 25°C. Calculate the pKb for hypochlorous anions.
6.48 Ka and Kb are related by the equation: Ka × Kb = Kw = 1.0*10^-14. The relationship between pKa and pKb is as follows: pKa + pKb = pKw = 14.00. See Section 16.8 (There are a couple ways using these equations to get the answer. You can find Kb first and then find pKb or you can find pKa and then find pKb.)
The concentration of OH− in an aqueous solution is 8.0 × 10^−5 at room temperature. What is the pH of this solution?
9.90 [H3O+] = 1.2 × 10^−10 and pH = −log(1.2 × 10^−10) = 9.90. (I found pOH and subtracted it from 14, but Mastering seems to have used Kw = [H+][OH-]. In either case it gives the same answer.)
At 25°C, the following data were collected: Solution A: pH = 8.46 Solution B: pOH = 9.30 Solution C: [H+] = 2.5 × 10-4 M Solution D: [OH-] = 1.1 × 10-10 M Which lists the solutions in order of increasing acidity?
A < B < D < C (A: gives the pH already B: pH + pOH = 14.00 C: pH = -log[H+] D: Either pOH = -log[OH-] then use the equation for B or [H+][OH-] = 1.0*10^-14 to find [H+] then use the equation for C.)
How many of the following salts will form acidic solutions when dissolved in water? AlCl3 Na2SO3 CH3NH3NO3?
AlCl3 and CH3NH3NO3 AlCl3 will form an acidic solution when it dissolves due to the reaction between Al3+ and water. When CH3NH3NO3 dissolves, the NO3 - anion will not react with water, but the CH3NH3 + ion, being a conjugate of a weak base, will react with water, CH3NH3+(aq) + H2O(l) ⇌ CH3NH2(aq) + H3O+(aq), turning the solution acidic.
A Lewis base is defined to be An electron-pair acceptor. A proton donor. An electron-pair donor. A proton acceptor.
An electron-pair donor.
Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs. AsO3^3− or AsO4^3−.
AsO3^3− (My guess for how to determine the correct answer is: H3AsO3 and H3AsO4 are both oxyacids. The strength of the oxyacid increases with each additional electronegative atom bonded to the central atom so H3AsO3 should be less acidic than H3AsO4 because it has one fewer oxygen atom. Because it is the weaker acid, it would be the stronger base.)
The equilibrium-constant of the reaction NO2(g)+NO3(g)⇌N2O5(g) is K=2.1×10^−20. What can be said about this reaction? At equilibrium the concentration of products and reactants is about the same. At equilibrium the concentration of products is much greater than the concentration of reactants. At equilibrium the concentration of reactants is much greater than that of products. There are no reactants left over once the reaction reaches equilibrium.
At equilibrium the concentration of reactants is much greater than that of products. Consider how the numerical value of K is influenced by the magnitude of the concentrations of the reactants and the products. For instance, if the concentrations of products are relatively large at equilibrium compared to the concentrations of reactants, the numerator of the equilibrium-constant expression will be quite large compared to the denominator, and the value of K will be correspondingly large. By analyzing the equilibrium-constant expression we can determine that a larger concentration of products in the numerator of the expression will result in larger values for K. Greater values of K are associated with reactions running toward completion, whereas smaller values of K are associated with a larger concentration of reactants. K⋘1 Reaction favors reactants, Reaction lies to left K∼1 Reaction favors neither reactants nor products, Reaction lies to center K⋙1 Reaction favors products, Reaction lies to right Note that reaction stoichiometries with a large difference between the number of moles of reactants and products may not strictly follow these general guidelines.
Which, if any, of the following statements are true? A. The stronger the base, the larger the pKb. B. The stronger the base, the larger the pKa of its conjugate acid. C. The stronger the base, the larger the Kb. D. The stronger the base, the smaller the Kb. E. The stronger the base, the smaller the pKa of its conjugate acid. F. The stronger the base, the smaller the pKb.
B. The stronger the base, the larger the pKa of its conjugate acid. C. The stronger the base, the larger the Kb. F. The stronger the base, the smaller the pKb. (As pKa/pKb gets larger, the weaker the acid/base respectively. Whereas as Ka/Kb gets larger, the stronger the acid/base respectively. Because Strong acids mean weak conjugate bases and strong bases mean weak conjugate acids if Kb is large then Ka of the conjugate acid is small, but the pKa is large.)
https://session.masteringchemistry.com/problemAsset/3238288/1/Brown13.ch16.f16.jpg Why do we need to use multiple acid-base indicators in this figure? Because the range of pH values is too large. Because the given salts dissolve in water forming colored solutions. Because Zn2+ and Al3+ ions react with some indicators. Because the salt concentrations are different.
Because the range of pH values is too large.
Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs. BrO− or ClO−.
BrO− (Compare the Ka values of their conjugate acids. The weaker conjugate acid will be the stronger base.)
What is the conjugate base of HCO3^−?
CO3^2− The conjugate base will have one less proton.
Which of the following is a weak base? ClO- HNO3 NH4+ KOH
ClO- Weak bases fall into two general categories: (1) neutral substances that have an atom with a nonbonding pair of electrons that can accept a proton and (2) anions of weak acids. See Section 16.7 (HNO3 and NH4+ are both acids and KOH is a strong base. ClO- is the conjugate of HClO which is a weak acid. Therefore ClO- is a weak base.)
Which equilibrium reaction will experience a shift towards the products in equilibrium position when the concentration of Ni2+ is increased? A. [Ni(H2O)6]2+(aq) + 6NH3(aq) ⇌ [Ni(NH3)6]2+(aq) + 6H2O(l) B. NiS(s) ⇌ Ni2+(aq) + S2−(aq) C. Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH−(aq) D. Ni2+(aq) + 6NH3(aq) ⇌ [Ni(NH3)6]2+
D. Ni2+(aq) + 6NH3(aq) ⇌ [Ni(NH3)6]2+ When the concentration of a reactant increases, the system will respond by relieving the stress on the system and moving towards the production of more product. When the concentration of a product increases, again the system will respond by relieving the stress and moving towards the production of reactant
True of False In a series of acids that have the same central atom, acid strength increases with the number of hydrogen atoms bonded to the central atom.
False (Acid strength increases as additional *electronegative* atoms bonds to the central atom.)
True or False Hydrotelluric acid (H2Te) is a stronger acid than H2S because Te is more electronegative than S.
False (H2Te is a stronger acid, but Te is *less* electronegative than S. H2Te is a stronger acid because the bond strength between Te and H is weaker than between S and H. Acidity increases as bond strength decreases.)
The following equation shows the equilibrium in an aqueous solution of methylamine: CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq) Which of the following represents a conjugate acid-base pair? CH3NH2 and H2O CH3NH3+ and OH− H2O and OH− CH3NH2 and OH−
H2O and OH− In the presence of methylamine (a weak base), H2O acts as an acid and donates a proton. OH− is a conjugate base of H2O. The simplest way to identify a conjugate acid-base pair is to find the species whose formulas differ only by one proton. According to the Brønsted-Lowry theory, an acid is any substance (molecule or ion) that can transfer a proton (H+ ion) to another substance, and a base is any substance that can accept a proton. Acid-base reactions are proton-transfer reactions, as follows: HA + B ⇌ BH+ + A− acid base acid base Chemical species whose formulas differ only by one proton are said to be conjugate acid-base pairs. Thus, A− is the conjugate base of the acid HA, and HA is the conjugate acid of the base A−. Similarly, B is the conjugate base of the acid BH+, and BH+ is the conjugate acid of the base B. The stronger the acid, the weaker the conjugate base, and the stronger the base, the weaker the conjugate acid.
Write a chemical equation that illustrates the autoionization of water. Express your answer as a chemical equation. Identify all of the phases in your answer.
H2O(l)+H2O(l)⇌OH−(aq)+H3O+(aq) (H2O(l)⇌OH−(aq)+H+(aq) is probably also okay) Water is an amphiprotic compound, which means it can act as either an acid or a base. It can accept a proton, becoming H3O+, or donate a proton, becoming OH−.
What is the conjugate acid of HPO3^2−?
H2PO3^− The conjugate acid will have one more proton.
Which correctly lists the acids in increasing order of strength? HBrO4 < HBrO2 < HBrO3 < HBrO HBrO < HBrO2 < HBrO3 < HBrO4 HBrO3 < HBrO2 < HBrO4 < HBrO HBrO4 < HBrO3 < HBrO2 < HBrO
HBrO < HBrO2 < HBrO3 < HBrO4 The strength of an oxyacid increases as additional electronegative atoms bind to the central atom. See Section 16.10
Which of the following actions will cause the equilibrium to shift such that the concentration of NO gas will increase? 2NO(g) + 2H2(g) ⇌ N2(g)+2H2O(g)+heat Increase the concentration of N2(g) Lower the reaction temperature Increase the concentration of H2(g) Decrease the concentration of N2(g)
Increase the concentration of N2(g) When the concentration of a reactant increases, the system will respond by relieving the stress on the system and moving towards the production of more product. When the concentration of a product increases, again the system will respond by relieving the stress and moving towards the production of reactant. Energy can be though of as a reagent or product in this sense. With an exothermic reaction, an increase in heat will shift the reaction towards the reverse reaction.
The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles in liquid water according to the equation 2HgO(s) + H2O(l) + 2Cl2(g) ⇌ 2HOCl(aq) + HgO⋅HgCl2(s) What is the equilibrium-constant expression for this reaction?
K = [HOCl]^2 / [Cl2]^2 Heterogeneous equilibria involve reactants and products that are not all in the same phase. For example, a solid may decompose, forming two gases. When writing equilibrium-constant expressions for heterogeneous reactions, generally the concentrations of pure solids and liquids are omitted. The composition of a pure solid or liquid does not change over the course of the reaction; only its quantity changes. Since its concentration isn't changing, the values can usually be excluded. Given the equation xCD(s)⇌cC(g)+dD(g), the equilibrium constant expression would be: K=[C]^c / [D]^d
Identify the proper form of the equilibrium-constant expression for the equation N2(g)+O2(g)⇌2NO(g)
K = [NO]^2 / [N2][O2] The equilibrium-constant expression is used to describe the concentration of reactants and products for a reaction in dynamic equilibrium. For ideal gases and ideal solutions in homogeneous equilibria, where all reactants and products are in the same phase, the extent to which a particular chemical reaction proceeds to products is given by the equilibrium equation: aA+bB⇌cC+dD, K = [C]^c[D]^d / [A]^a[B]^b where K is the equilibrium constant and the right-hand side of the equation is known as the equilibrium-constant expression. The concentration of each product raised to its coefficient is divided by the concentration of each reagent raised to its coefficient according the the balanced chemical equation. Therefore, the higher the concentration of products, the larger the value of K will be.
Calculate Ka value for NH4+. Express your answer using two significant figures. (Kb for ammonia is 1.8×10^−5)
Ka = 5.6×10^−10 (Ka x Kb = Kw = 1.0 x 10^-14)
Benzoic acid C6H5COOH is a commonly used antimicrobial preservative in food and beverages, especially in carbonated beverages. A 0.100 M solution of C6H5COOH has a pH of 2.60. Calculate the Ka value for this acid. Express the equilibrium constant to two significant figures.
Ka = 6.5×10^−5 C6H5COOH dissociates into H+ and C6H5COO−. First, determine [H+] using the given pH, 2.60, in the following equation: pH=−log[H+] Rearrange and cancel out the log to solve for [H+]: [H+] = antilog(−pH) = 10^−pH = 10^−2.60 = 2.5×10^−3 M Because each mole of C6H5COOH dissociates into one mole of H+ and one mole of C6H5COO−, C6H5COO− will have the same molarity as H+. By tabulating the data for initial, change, and equilibrium concentrations, you can find that the equilibrium concentration of C6H5COOH is calculated by subtracting the concentration of H+ from the total concentration of the solution, 0.100 M: ___________ C6H5COOH ___________ H+__________ C6H5COO− IC (M) _____ 0.100 M ______________ 0 M __________ 0 M CC (M) ____ −2.5E−3 M ________ +2.5E−3 M ____ +2.5E−3 M EC (M) ___ 0.100 − 2.5E−3 M _____ 2.5E−3 M ____ 2.5E−3 M Thus, the equilibrium concentration of C6H5COOH is [C6H5COOH] = 0.100 M − 2.5E−3 M = 9.75E−2 M Finally, substitute this value into the Ka expression and solve for Ka: Ka = [H+][A−] / [HA] = [H+][C6H5COO−] / [C6H5COOH] = (2.5E−3)^2 / 9.75E−2 = 6.5E−5
Calculate Ka value for H3NOH+. Express your answer using two significant figures. (The Kb for hydroxylamine is 1.1×10^−8.)
Ka = 9.1×10^−7 (Ka x Kb = Kw = 1.0x10^-14)
For the reaction X(g)+2Y(g)⇌2Z(g) Kp = 1.78×10^−2 at a temperature of 255 ∘C. Calculate the value of Kc.
Kc = 0.771 The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)^Δn where R=0.08206 L⋅atm/(K⋅mol), T is the absolute temperature, and Δn is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction N2(g)+3H2(g)⇌2NH3(g) for which Δn=2−(1+3)=−2.
Methanol (CH3OH) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: CO(g)+2H2(g)⇌CH3OH(g). An equilibrium mixture in a 2.50 L vessel is found to contain 0.0262 mol CH3OH, 0.170 mol CO, and 0.303 mol H2 at high temperature. Calculate Kc at this temperature. Express your answer to three significant figures.
Kc = 10.5 To solve the problem, begin by writing the expression for the equilibrium constant. The equilibrium constant is the ratio between the product of the concentrations of the reactants raised to their stoichiometric coefficients and the product of the concentrations of the products raised to their stoichiometric coefficients. Thus, the expression for the equilibrium constant is as follows: Kc=[CH3OH][CO][H2]2 Now calculate the equilibrium concentrations of each reactant using their equilibrium number of moles and the volume of the vessel: [CH3OH] = 0.0262 mol CH3OH / 2.50L = 1.05×10^−2M [CO] = 0.170 mol CO / 2.50L = 6.80×10^−2M [H2] = 0.303 mol H2 / 2.50L = 0.121M Substitute the obtained concentrations into the above equation to find Kc: Kc = 1.05×10^−2 / (6.80×10^−2 × 0.121^2) = 10.5
A mixture initially contains A, B, and C in the following concentrations: [A] = 0.500 M , [B] = 1.00 M , and [C] = 0.300 M. The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.370 M and [C] = 0.430 M . Calculate the value of the equilibrium constant, Kc.
Kc = 2.12 The Kc expression for the reaction A+2B⇌C is: Kc = [C] / ([A][B]^2) To determine how the concentrations change, it is useful to set up a table below the balanced chemical equation that shows the initial concentrations, the change in concentration, and the equilibrium concentrations. ___________________A___________+2B__________⇌C Initial (M):_________ 0.500 ________ 1.00 _______ 0.300 Change (M):________ −x __________ −2x _________ +x Equilibrium (M):_____ 0.370_____________________ 0.430 What is the value of x? If x = 0.130, what is the final concentration of B?
A chemical reaction between X2 (red) and Y2 (blue) produces XY(red-blue). All species are in a gaseous state. The picture shown here represents the equilibrium mixture. https://session.masteringchemistry.com/problemAsset/1037081/42/MF_1037081_molecules.jpg Calculate the equilibrium constant for the balanced reaction between one mole of each reagent. Express your answer numerically to three significant figures.
Kc = 2.72 First, write the balanced chemical equation. Use the balanced equation to write the equilibrium-constant expression. Molarity is the number of moles divided by volume, and the number of moles is the number of molecules divided by Avogadro's number. Therefore, the molar concentrations in the Kc expression will be directly proportional to the number of molecules you see in the picture (three X2 molecules, six Y2 molecules, and seven XY molecules). The balanced equation for this reaction is: X2+Y2⇌2XY Then do Kc = [XY]^2 / [X2][Y2] When written using the lowest possible whole-number coefficients, 1 mol of each reagent reacts according to the balanced equation X2+Y2⇌2XY You could also think of this reaction in terms of the product forming (1/2)X2 + (1/2)Y2 ⇌ XY The calculated value for Ka when written in terms of 1 mol of XY is equivelent to √(2.72) = 1.65.
A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00 L vessel at 700 K. These substances react as follows: H2(g)+Br2(g)⇌2HBr(g) At equilibrium the vessel is found to contain 0.566 g of H2. Calculate Kc. Express your answer to two significant figures.
Kc = 58 Plugging the equilibrium concentrations into the equilibrium expression for this reaction yields the equilibrium constant. The coefficients from the balanced chemical equation become the exponents in the equilibrium expression for Kc. The calculation of Kc is summarized below: Kc = [HBr]^2 / [H2][Br2] = (0.4008)^2 / ((0.1404)(1.959×10^−2)) = 58 For reference the ICE table is as follows: _____nH2(mol) ___ *[H2](M)* ___ nBr2(mol) ___ *[Br2](M)* ___ nHBr(mol) ___ *[HBr](M)* Initial: _ 0.6815 ____ *0.3408* ____ 0.440 _______ *0.220* ______ 0 __________ *0 M* --------------------------------------------------------------------------------------------------------------------- Change: −0.4008 __ *−0.2004* __ −0.4008 ____ *−0.2004* ____ +2(0.4008) ___ +2(0.2004) -------------------------------------------------------------------------------------------------------------------- Equilibrium: 0.281 ____ *0.140* _____ 0.039 _____ *0.020* _____ 0.802 _______ 0.401 ---------------------------------------------------------------------------------------------------------------------
What is the correct equilibrium constant expression for the following heterogeneous equilibrium? 4HCl (g) + O2 (g) ⇄ 2H2O (l) + 2Cl2 (g)?
Kc = [Cl2]^2 / [HCl]^4[O2] (Liquids and Solids do not appear in the expression)
Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.89×10^−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.24×10^−4, what is the equilibrium constant Kfinal for the following reaction? PbCl2(aq) + 2Ag+ (aq) ⇌ 2AgCl(aq) + Pb2+(aq)
Kfinal = 1.23×10^−2 When a chemical equation is the sum of two chemical equations for which equilibrium constants are already known, the equilibrium constant for the reaction is the product of the equilibrium constants for the individual reactions (K3 = K1 * K2). When a reaction is written in reverse, the new equilibrium constant for the reversed equation is the inverse of the original equilibrium constant, K' = 1/K. When the stoichiometry of a reaction changes, the new equilibrium constant is raised to the same power. Thus, if a reaction is multiplied by 2, the equilibrium constant K is squared. (Notice that in order to get the final equation the K4 equation must be doubled and reversed. Because the equation is doubled the new coefficient becomes the power of the concentrations in the equations. This means to solve the equation you must both square the value for K4 and get its reciprocal (in either order is okay). Then take that value and multiply it by the K3 value to get the answer.)
Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.52×10^−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.07×10^−19, what is the equilibrium constant Kfinal for the following reaction? S2−(aq)+2H+(aq)⇌H2S(aq)
Kfinal = 9.82×10^25 (Notice that Kfinal is for the flipped equation which means you need to take the reciprocal of the answer for K3 below to get Kfinal.) For a chemical reaction equation with the general form aA+bB⇌cC+dD the equilibrium equation is given by K1=[C]^c[D]^d / [A]^a[B]^b Thus, for a chemical reaction equation with the general form cC+dD⇌eE+fF the equilibrium equation is given by K2=[E]^e[F]^f / [C]^c[D]^d If the first two equations are added together such that aA+bB⇌eE+fF then the equilibrium equation is given by K3=[E]^e[F]^f / [A]^a[B]^b = [C]^c[D]^d / [A]^a[B]^b ⋅ [E]^e[F]^f / [C]^c[D]^d or K3=K1⋅K2 Thus, when a chemical equation is the sum of two chemical equations for which equilibrium constants are already known, the equilibrium constant for the reaction is the product of the equilibrium constants for the individual reactions. Keep in mind that equilibrium equations do not include expressions for any pure solids or liquids that may be involved in the reaction.
For the reaction 3A(g)+3B(g)⇌C(g) Kc = 55.4 at a temperature of 395 ∘C. Calculate the value of Kp.
Kp = 1.12×10^−7 The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)^Δn where R=0.08206 L⋅atm/(K⋅mol), T is the absolute temperature, and Δn is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction N2(g)+3H2(g)⇌2NH3(g) for which Δn=2−(1+3)=−2.
Write the expression for the ion-product constant for water, Kw.
Kw=[H+][OH−] The ion-product constant for water excludes the term [H2O] from the equilibrium-constant expression since it is a pure liquid. Kw is equal to 1.0×10−14 at 25 ∘C.
For the decomposition of A to B and C A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibrium? (Leftward shift, No shift, Rightward shift) A. Double the concentrations of both products. B. Double the concentrations of both products and then double the container volume. C. Double the concentrations of both products and then quadruple the container volume. D. Double the concentration of B and halve the concentration of C. E. Add more A. F. Double the container volume.
Leftward shift: A. Double the concentrations of both products. No shift: B. Double the concentrations of both products and then *DOUBLE* the container volume. D. *Double* the concentration of *B* and *halve* the concentration of *C*. E. Add more A. Rightward shift: C. Double the concentrations of both products and then *QUADRUPLE* the container volume. F. Double the container volume.
For the following systems at equilibrium C: CaCO3(s) ⇌ CaO(s) + CO2(g), ΔH = +178 kJ/mol D: PCl3(g) + Cl2(g) ⇌ PCl5(g), ΔH = −88 kJ/mol classify these changes by their effect. (Based on any shift that will occur) System C Increase temperature System C Decrease temperature System D Increase temperature System D Decrease temperature
Leftward shift: System *C Decrease* temperature System *D Increase* temperature No shift: (None) Rightward shift: System *C Increase* temperature System *D Decrease* temperature There are two ways to approach this problem. The first method involves assigning heat as a reactant or product, and then treating changes in temperature the same way you would an increase in reactant or product concentration. An endothermic reaction (positive ΔH) absorbs heat input, whereas an exothermic reaction (negative ΔH) releases heat. Endothermic: heat+reactants ⇌ products Exothermic: reactants ⇌ products+heat Another approach uses the fact that: K is directly proportional to temperature for endothermic reactions, and K is inversely proportional to temperature for exothermic reactions. Once you know how a particular temperature change affects the value of K, you'll be able to tell whether Q is greater than or less than K and, thus, how the reaction will respond.
For the following systems at equilibrium A:2NOCl(g)⇌2NO(g)+Cl2(g) B:H2(g)+I2(g)⇌2HI(g) classify these changes by their effect. (Based on any shift that will occur) System A Increase container size System A Decrease container size System B Increase container size System B Decrease container size
Leftward shift: System A Decrease container size No shift: System B Decrease container size System B Increase container size Rightward shift: System A Increase container size Note that the equilibrium of a gaseous reaction is only disturbed by a change in volume if the two sides of the chemical reaction have different numbers of moles of gas. If the sum of the coefficients of the gaseous species is the same on both sides of the equation, then a volume change has no effect on equilibrium because the resulting pressure and concentration differences cancel out in the reaction quotient.
For the reaction C(s)+H2O(g)⇌H2(g)+CO(g) classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the reaction. add some solid carbon remove some hydrogen add some carbon monoxide remove some water remove some solid carbon add some hydrogen remove some carbon monoxide add some water
Leftward shift: add some carbon monoxide remove some water add some hydrogen Rightward shift: remove some hydrogen remove some carbon monoxide add some water No shift: add some solid carbon remove some solid carbon Start by determining whether the component of the reaction that is changed is a reactant or product. From this, determine how this change affects the reaction quotient, Q. K (and Q) = [H2][CO] / [H2O] (Note that the solid carbon is not included in the expression which is why varying its amount doesn't shift the reaction in either direction.)
For the reaction N2(g)+O2(g)⇌2NO(g) classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the reaction. double [oxygen] halve [nitrogen monoxide] double [nitrogen] double [nitrogen monoxide] halve [nitrogen] halve [oxygen]
Leftward shift: double [nitrogen monoxide] halve [oxygen] halve [nitrogen] Rightward shift: double [oxygen] halve [nitrogen monoxide] double [nitrogen] No shift: (None) Start by determining whether the component of the reactant that is changed is a reactant or product. From this, determine how this change affects the reaction quotient, Q. Finally, compare the new Q to the equilibrium constant, K, of the reaction to determine in which direction, if any, the reaction will respond. If Q>K, the reaction will shift toward the reactants. If Q<K, the reaction will shift toward the products.
For the reaction of A and B forming C, A(g)+B(s)⇌2C(g) how will the reaction respond to each of the following changes at equilibrium? halve the concentration of C quadruple the number of moles of B halve the concentration of A double the concentrations of A and C double the concentration of A
Leftward shift: halve the concentration of A double the concentrations of A and C No shift: quadruple the number of moles of B Rightward shift: halve the concentration of C double the concentration of A (Doubling the concentrations of A and C results in A^2 and C^4 which is why it shifts to the left.)
Which of the following salts, when added to water, will result in a solution with pH = 7.00? Ca(NO2)2 NH4NO3 NaNO3 FeCl3
NaNO3 A neutral solution will result when a salt, whose anion is the conjugate base of a strong acid and whose cation is either from group 1A or one of the heavier members of group 2A (Ca2+, Sr2+, or Ba2+), is added to water. See Section 16.9 (NO3- is the conjugate base of HNO3 which is a strong acid and therefore has negligible basicity. Na+ is a cation from group 1A. Because of this, NaNO3 does not change pH. For Ca(NO2)2, NO2- is the conjugate base of a weak acid and therefore is a weak base. Ca2+ is a heavier member of group 2A so as mentioned above will not alter pH. This salt will make the solution basic. For NH4NO3, NO3- is the conjugate base of HNO3 which is a strong acid and therefore has negligible basicity. NH4+ is the conjugate acid of NH3 which is a weak base making NH4+ a weak acid. This salt will make the solution acidic. For FeCl3, Fe3+ is a small cation with a charge of 2+ or greater which means it will act as a weak acid (it will attract unshared electrons on the oxygen in water molecules, weaken the O-H bonds and some H get transferred to solvent water molecules to create H3O+). Cl- is the conjugate base of HCl which is a strong acid and therefore will have negligible basicity. Overall this solution will be acidic.)
Classify each salt as acidic, basic, or neutral. CH3NH3NO3 BaBr2 NaNO2
Neutral salts: BaBr2 Acidic salts: CH3NH3NO3 Basic salts: NaNO2 (BaBr2: Br- is the conjugate base of HBr which is a strong acid therefore Br- has negligible basicity. Ba2+ is a heavier element from Group 2A and so has negligible acidity. Overall this salt is neutral. CH3NH3NO3: NO3- is the conjugate base of HNO3 which is a strong acid therefore NO3- has negligible basicity. CH3NH3+ is the conjugate acid of methylamine which is a weak base therefore CH3NH3+ is a weak acid. Therefore this is an acidic salt. NaNO2: NO2- is the conjugate base of HNO2 which is a weak acid therefore NO2- is a weak base. Na is a Group 1A element so it has negligible acidity. Overall NaNO2 will be basic) Salts that are derived from the reaction of a weak acid with a strong base consist of a spectator ion (from the strong base) and an anion that is a relatively strong conjugate base. The conjugate base will hydrolyze to form OH− ions, giving the solution a basic character. Similarly, salts that are derived from the reaction of a weak base with a strong acid consist of a spectator ion (from the strong acid) and a cation that is relatively strong conjugate acid. The conjugate acid will hydrolyze to form H3O+ ions, giving the solution acidic character. A salt is an ionic compound that is produced when a cation and an anion from an acid-base reaction combine. In other words, the cation from a base replaces a proton on an acid. An example is the reaction of the strong base NaOH with the strong acid HCl. In solution, the H+ and OH− ions react to form H2O, leaving a solution of salt ions, Na+ and Cl−. In this example, the salt ions do not affect the pH of the solution and are, therefore, said to be spectator ions. The conjugate base of a strong acid is always a weak base (and the conjugate acid of a strong base is always a weak acid). So the conjugate base Cl− of the strong acid HCl is always a weak base and the conjugate acid Na+ of a strong base NaOH is always a weak acid. However, anions from weak acids and cations from weak bases change the pH of an aqueous solution. If one of the salt ions is the conjugate acid of a weak base, the ion will react with H2O to create H3O+ ions, acidifying the solution. Similarly, the conjugate base of a weak acid will react with H2O to create OH− ions, making the solution more alkaline.
https://session.masteringchemistry.com/problemAsset/3237967/1/Brown13_15-10.jpg Effect of adding H2 to an equilibrium mixture of N2, H2, and NH3. Adding H2 causes the reaction as written to shift to the right, consuming some H2 to produce more NH3. Why does the nitrogen concentration decrease after hydrogen is added? N2(g)+3H2(g)⇌2NH3(g) Nitrogen is converted into the solid state. Nitrogen (and some of the added hydrogen) is converted into ammonia. Nitrogen is converted into hydrogen. Nitrogen is converted into the liquid state.
Nitrogen (and some of the added hydrogen) is converted into ammonia.
Would the line go exactly through the origin of the plot? https://session.masteringchemistry.com/problemAsset/3238316/1/Brown13.ch16.p.06.jpg
No. (The autoionization of water would mean [H+] will always be greater than zero even when the concentration of the acid is zero and so the line will not go exactly through the origin.)
At 100 ∘C, Kc = 0.078 for the reaction SO2Cl2(g) ⇋ SO2(g) + Cl2(g). In an equilibrium mixture of three gases, the concentrations of SO2Cl2 and SO2 are 0.142 M and 0.143 M respectively. What is the partial pressure of Cl2 in the equilibrium mixture? Express your in atmospheres to two significant figures.
P = 2.4atm (Use Kc to find [Cl2] then use P = MRT to find the partial pressure).
Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs. S^2− or HS−.
S2− (Both S^2- and HS- come from H2S which is a weak acid. HS- will be a stronger acid than S^2- so S^2- will be the stronger base.)
The graph given below (Figure 1) shows [H+] vs. concentration for an aqueous solution of an unknown substance. https://session.masteringchemistry.com/problemAsset/3238316/1/Brown13.ch16.p.06.jpg Is the substance a strong acid, a weak acid, a strong base, or a weak base?
Strong acid
Among three bases, X−, Y−, and Z−, the strongest one is Y−, and the weakest one is Z−. Rank their conjugate acids, HX, HY, and HZ, in order of decreasing strength. Rank the acids from strongest to weakest.
Strongest acid: HZ Middle: HX Weakest acid: HY
Consider the exothermic reaction Fe2O3(s)+2Al(s)⇌Al2O3(s)+2Fe(s) In what direction will the equilibrium shift when heat is removed?
The equilibrium will shift toward the products. For an exothermic reaction, heat is a product in the reaction. Consider what effect the removal of heat will have on the overall reaction.
Considering the reaction below at equilibrium, which of the following statements is correct? N2O4 (g) ⇄ 2NO2 (g) The concentration of N2O4 is equal to the concentration of NO2. N2O4 can no longer react to form NO2. The ratio of [NO2]2/[N2O4] is not changing. NO2 is being formed more quickly than N2O4.
The ratio of [NO2]2/[N2O4] is not changing. Chemical equilibrium occurs when opposing reactions proceed at equal rates.
When this reaction in equilibrium is heated, a red/brown color persists. What is true about this reaction? 2NO2(g) ⇋ N2O4(g) brown/red colorless The increase in T results in a high K value. ΔH for this reaction is positive. Heat acts as a reactant in this equilibrium. The reaction is exothermic.
The reaction is exothermic. When heat is added a red/brown color persists so this reaction equilibrium has shifted towards the reactant side. Thus, heat can be thought of as a product in this exothermic equilibrium.
Now consider the reaction A+2B⇌C for which in the initial mixture Qc = [C] / [A][B]^2 = 387 Is the reaction at equilibrium? If not, in which direction will it proceed to reach equilibrium? (Note: The previously calculated value for Kc was 2.12)
The reaction will proceed in reverse to form reactants.
If the reaction N2(g)+3H2(g)⇌2NH3(g) is at equilibrium, what direction will the reaction shift if NH3 gas is added?
The reaction will shift to produce reactants. Think about Le Châtelier's principle and how the reaction will need to shift to overcome the added stress.
If an endothermic reaction is at equilibrium, how will the equilibrium shift when heat is added to the reaction?
The reaction will shift toward the products. For an endothermic reaction, heat is a reactant in the reaction. Think about what effect the addition of heat will have on the overall reaction.
Consider the reaction 2H2(g)+O2(g)⇌2H2O(g) How will the equilibrium shift when hydrogen gas is removed?
The system will shift toward the reactants. Think about Le Châtelier's principle and how the reaction will need to shift to overcome the added stress.
The following pictures represent the equilibrium mixtures of five different chemical reactions. All the reactions proceed according to the general balanced chemical reaction A+B⇌AB Red spheres represent A atoms, blue spheres are B atoms, and red-blue clusters are AB molecules. Rank the reactions in order of increasing equilibrium constant.
There are a couple ways to do this problem: Method 1: Remember that the larger the value of K the more the reaction leans toward the products. As such, you can count the number of AB molecules. The more there are the larger the equilibrium constant. This means you can just order the pictures from least number of AB molecules to the greatest number. Method 2: Use the following formula to actually find what the values of K are: Kc = [AB] / [A][B] where the concentrations are simply the number of particles of each (e.g. [AB] = 3 while [A] and [B] are both 7 so do 3 / 49 to find K).
Where does the equilibrium of this reaction lie? (The equilibrium-constant of the reaction NO2(g)+NO3(g)⇌N2O5(g) is K=2.1×10^−20.)
To the left Recall that if products are favored, the equilibrium is said to lie to the right, since more of the chemical species from the right-hand side of the equation are at equilibrium. Similarly, if reactants are favored, the equilibrium is said to lie to the left, since more of the chemical species from the left-hand side of the equation will be present at equilibrium.
Indicate whether each of the following statements is true or false. In general, the acidity of binary acids increases from left to right in a given row of the periodic table.
True
The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc = [C][D] / [A][B] = 5.0 Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached? Express your answer to two significant figures and include the appropriate units.
[A] = 0.62 M (Create an ICE table and fill in blanks with variables. Plug variables into equilibrium expression which will give a quadratic equation. Solve for x then find [A] by subtracting x from 2.)
A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00 L vessel at 700 K. These substances react as follows: H2(g)+Br2(g)⇌2HBr(g) At equilibrium the vessel is found to contain 0.566 g of H2. Calculate the equilibrium concentration of Br2. Express the molarity to two significant figures.
[Br2]eq = 2.0×10^−2M The amounts of Br2 and HBr at equilibrium are derived using the stoichiometric relationship between H2 and each species. The table showing the moles and concentrations of the reactants of the reaction are summarized below. nH2(mol) _ *[H2](M)* _ nBr2(mol) _ *[Br2](M)* _ nHBr(mol) _ *[HBr](M)* I: 0.6815 __ *0.3408* ___ 0.440 ____ *0.220* ______ 0 ______ *0 M* -------------------------------------------------------------------------------------------- C:−0.4008 _ *−0.2004* _ −0.4008 _ *−0.2004* -------------------------------------------------------------------------------------------- E: _ 0.281 __ *0.140* ____ 0.039 ___ *0.020* --------------------------------------------------------------------------------------------
Many household cleaning products contain oxalic acid, H2C2O4, a diprotic acid with the following dissociation constants: Ka1 = 5.9×10^−2, Ka2 = 6.4×10^−5. Calculate the equilibrium concentration of C2O42− in a 0.20 M solution of oxalic acid. Express your answer to two significant figures and include the appropriate units.
[C2O42−] = 6.4×10^−5 M (You'll need to do 2 sets of ICE tables. The answer is the y variable from the 2nd ICE table.)
The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc = [C][D] / [A][B] = 5.0 What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ? Express your answer to two significant figures and include the appropriate units.
[D] = 0.87 M (Create an ICE table and fill in blanks with info and variables. Plug info and variables into equilibrium expression which will give a quadratic equation. Solve for x.)
If a solution is described as basic, which of the following is true? [H+]>[OH−] [H+]<[OH−] [H+]=[OH−]
[H+]<[OH−] A base is a substance that increases the concentration of OH− ions when dissolved in water. Therefore, in a basic solution the concentration of OH− will be greater than the concentration of H+.
If a neutral solution of water, with pH=7.00, is cooled to 10 ∘C, the pH rises to 7.27. Which of the following three statements is correct for the cooled water? [H+]>[OH−] [H+]=[OH−] [H+]<[OH−]
[H+]=[OH−] In pure water, the only source of H+ is the autoionization reaction, which produces equal concentrations of H+ and OH−. As the temperature of water changes, the value of the ion-product constant for water (Kw) changes, and the pH at which [H+]=[OH−] changes. At 25 ∘C, the value of Kw for H2O is 1×10^−14. This gives [H+] = [OH−] = 1×10^−7, which is why a neutral solution of pure water at 25 ∘C has a pH = 7.00. It is [H+] = [OH−] that defines the neutral solution and not a pH equal to 7.00. Changing the temperature will change the value of Kw and, at temperatures different than 25 ∘C, a pH of 7.00 will not be a neutral solution.
A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00 L vessel at 700 K. These substances react as follows: H2(g)+Br2(g)⇌2HBr(g) At equilibrium the vessel is found to contain 0.566 g of H2. Calculate the equilibrium concentration of H2. Express the molarity to three significant figures.
[H2]eq = 0.140M The mass of H2 found at equilibrium is given, so the molarity of H2 at equilibrium can be directly calculated using molar mass and volume. [H2]eq = (0.566 g H2 / 2.00 L) × (1 mol / 2.016 g H2) = 0.1404M
Many household cleaning products contain oxalic acid, H2C2O4, a diprotic acid with the following dissociation constants: Ka1 = 5.9×10^−2, Ka2 = 6.4×10^−5. Calculate the equilibrium concentration of H3O+ in a 0.20 M solution of oxalic acid. Express your answer to two significant figures and include the appropriate units.
[H3O+] = 0.083 M (You'll need to do 2 sets of ICE tables. Well in this case not really because the answer to 2 sig figs doesn't actually change, but you're supposed to do 2 ICE tables. 0.20 M is the initial concentration of the acid.) Polyprotic acids contain more than one dissociable proton. Each dissociation step has its own acid-dissociation constant, Ka1, Ka2, etc. For example, a diprotic acid H2A reacts as follows: H2A(aq)+H2O(l)⇌H3O+(aq)+HA−(aq) Ka1 = [H3O+][HA−] / [H2A] HA−(aq)+H2O(l)⇌H3O+(aq)+A2−(aq) Ka2 = [H3O+][A2−] / [HA−] In general, Ka2 = [A2−] for a solution of a weak diprotic acid because [H3O+]≈[HA−].
Calculate the concentration of H3O+ of a solution if the concentration of OH- at 25°C is 3.8 × 10-5 M and determine if the solution is acidic or basic. [H3O+] = 2.6 × 10^-10 M; the solution is acidic. 2.6 × 10^18 M; the solution is basic. [H3O+] = 2.6 × 10^-10 M; the solution is basic. 3.8 × 10^-19 M; the solution is acidic.
[H3O+] = 2.6 × 10^-10 M; the solution is basic. The concentrations of H3O+ and OH- are related by Kw through the equation: Kw = [H3O+][OH-]. See Section 16.3 pH = -log [H30+]. A pH over 7 is basic, under 7 is acidic.
A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00 L vessel at 700 K. These substances react as follows: H2(g)+Br2(g)⇌2HBr(g) At equilibrium the vessel is found to contain 0.566 g of H2. Calculate the equilibrium concentration of HBr. Express the molarity to three significant figures.
[HBr]eq = 0.401M The ICE table below summarizes the progress of the equilibrium reaction. The amount of H2 consumed is the difference between the initial amount (given mass converted to moles) and equilibrium amount (given mass converted to moles). The change in moles of Br2 consumed is equal to the change in moles of H2 consumed based on the stoichiometric relationship, and there are two moles of HBr produced for every one mole of H2 consumed. With the initial and change rows completed, the remaining cells in the equilibrium row can be completed. _____nH2(mol) ___ *[H2](M)* ___ nBr2(mol) ___ *[Br2](M)* ___ nHBr(mol) ___ *[HBr](M)* Initial: _ 0.6815 ____ *0.3408* ____ 0.440 _______ *0.220* ______ 0 __________ *0 M* --------------------------------------------------------------------------------------------------------------------- Change: −0.4008 __ *−0.2004* __ −0.4008 ____ *−0.2004* ____ +2(0.4008) ___ +2(0.2004) -------------------------------------------------------------------------------------------------------------------- Equilibrium: 0.281 ____ *0.140* _____ 0.039 _____ *0.020* _____ 0.802 _______ 0.401 ---------------------------------------------------------------------------------------------------------------------
Many household cleaning products contain oxalic acid, H2C2O4, a diprotic acid with the following dissociation constants: Ka1 = 5.9×10^−2, Ka2 = 6.4×10^−5. Calculate the equilibrium concentration of HC2O4− in a 0.20 M solution of oxalic acid. Express your answer to two significant figures and include the appropriate units.
[HC2O4−] = 0.083 M (You need the final equilibrium concentration for HC2O4- from the 2nd ICE table. In this case it ends up being the same as the equilibrium from the 1st ICE table, but that probably won't always be the case so use the 2nd ICE table's equilibrium value.) Because Ka2 for oxalic acid is much smaller than Ka1, essentially all the HC2O4− produced in the first dissociation step stays in the solution. The concentration of HC2O4− that was consumed in the second reaction is negligible compared to the concentration produced in the first reaction.
Calculate the molar concentration of OH− ions in a 0.070 M solution of ethylamine (C2H5NH2; Kb=6.4×10^−4). Express your answer to two significant figures.
[OH−] = 6.4×10^−3 M (Use Kb and ICE table to find the value of x which is the value for [OH-].)
Given that Kb for ammonia is 1.8×10^−5 and that for hydroxylamine is 1.1×10^−8, which is the stronger base?
ammonia (The larger the value the stronger the base.)
Using the equation kf / kr = a constant, predict whether the equilibrium constant for the process is greater than 1 or less than 1. https://session.masteringchemistry.com/problemAsset/3237970/2/BLB10.15.1.jpg
greater than 1 Since kf is larger than kr, the constant produced must be larger than 1. (Remember that the magnitude for k is inversely related to the activation energy. When the Ea is large, then k is small.)
The Kb for ammonia is 1.8×10^−5 and that for hydroxylamine is 1.1×10^−8. Which is the stronger acid, the ammonium ion or the hydroxylammonium ion?
hydroxylammonium ion (The stronger acid is the conjugate of the weaker base.)
Based on the shown energy profile (Figure 1), predict whether kf>kr or kf<kr. https://session.masteringchemistry.com/problemAsset/3237970/2/BLB10.15.1.jpg
kf>kr Consider the Arrhenius equation, k=Ae−Ea/RT As the activation energy (Ea) increases, the value of the constant (k) decreases. Since the reverse reaction requires a higher input of energy, it will produce the lower k value.
https://session.masteringchemistry.com/problemAsset/3238284/1/indicator1.jpg https://session.masteringchemistry.com/problemAsset/3238284/1/indicator2.jpg https://session.masteringchemistry.com/problemAsset/3238284/1/indicator3.jpg Which of these indicators is best suited to distinguish between a solution that has a pH less than 5 and one that has a pH greater than 5?
methyl red
A solution of sodium acetate (NaCH3COO) has a pH of 9.57. The acid-dissociation constant for acetic acid is 1.8×10^−5. What is the molarity of the solution? Express your answer to two significant figures and include the appropriate units.
molarity = 2.5 M (If you got 4.0⋅10^−15M then you used Ka. You need to find Kb and pOH.) Since sodium acetate is a strong electrolyte, it completely dissociates when dissolved in water to form Na+ and CH3COO−. The acetate ion, being the conjugate base of a weak acid, reacts with water to produce acetic acid and hydroxide ions, thereby increasing the pH of the solution: CH3COO−(aq) + H2O(l) ⇌ CH3COOH(aq) + OH−(aq) The equilibrium-constant expression is Kb = [CH3COOH][OH−] / [CH3COO−] The product of the acid-dissociation constant for an acid (Ka) and the base-dissociation constant for its conjugate base (Kb) equals the ion-product constant for water (Kw=1.0×10^−14) at 25 ∘C. Therefore, for CH3COO−, Kb = 5.6×10^−10. Since the sum of pH and pOH is always equal to 14.00 at 25 ∘C, the pOH of the solution is 4.43. Rearrange the equation for pOH to calculate the concentration of OH−: [OH−] = 10^−pOH = 10^−4.43 = 3.7×10^−5M Tabulate the concentrations of the species involved in the equilibrium reaction, letting x to be the initial concentration of NaCH3COO and, therefore, CH3COO−: __________ [CH3COO−] _____________ [CH3COOH] __________ [OH−] IC (M) _______ x ________________________ 0 ____________________ 0 CC (M) ___ −3.7×10^−5 _____________ +3.7×10^−5 ___________ +3.7×10^−5 EC (M) ____ (x − 3.7×10^−5) _________ 3.7×10^−5 ____________ 3.7×10^−5 Substitute the equilibrium concentrations and the value of the base-dissociation constant into the Kb expression. Since Kb = 5.6×10^−10 is small, assume that 3.7×10^−5 is small compared to x; thus, x−3.7×10^−5 ≈ x: (3.7×10^−5)(3.7×10^−5) / x = 5.6×10^−10 Solving the equation gives x = 2.5M. Since a concentration of 3.7×10^−5M is much smaller than 5% of 2.5M (the initial CH3COO− concentration), the assumption is valid. Therefore, the initial concentration of CH3COO− and the molarity of the NaCH3COO solution is 2.5M.
Based on your answer to part A, determine the value of the pH of the solution when the concentration is 0.11 M? Express your answer using two decimal places. (Note: it's a strong acid)
pH = 0.96 (As it is a strong acid [H+] = [HA] and pH = -log[H+])
Calculate the pH of each of the following strong acid solutions. 20.0 mL of 1.70 M HCl diluted to 0.600 L. Express your answer to three decimal places.
pH = 1.247 Since HCl is a strong acid, it ionizes completely as per the following equation: HCl(aq)→H+(aq)+Cl−(aq) According to this equation, one mole of the acid produces one mole of H+ ions. Therefore, the concentration of H+ ions in the diluted solution numerically equals the concentration of HCl. Both concentrated and diluted solutions contain the same amount of HCl in moles. Thus, begin by finding the number of moles of HCl in the concentrated solution using the volume converted to liters and the molarity of the concentrated solution as a conversion factor: molHCl = 20.0mL soln × (1 L soln / 1000 mL soln) × (1.70 mol HCl / 1 L soln) = 3.40E−2 mol HCl When using the equation Mconc×Vconc=Mdil×Vdil (M1V1 = M2V2) where M represents either the concentrated (conc) or diluted (dil) concentration and V represents volume, the value 3.40E−2 mol is equal to Mconc×Vconc. Thus, find the molarity of HCl in the diluted solution by dividing the number of moles by the volume of the diluted solution: Molarity HCldil. = (mol HCl / L of soln) = (3.40E−2 mol HCl / 0.600L) = 5.67E−2 mol / L = 5.67E−2M Since the concentration of H+ ions in this solution numerically equals the concentration of HCl, substitute the concentration value into the equation for pH to get pH = −log[H+] = −log(5.67E−2) = 1.247
Calculate the pH of each of the following strong acid solutions. A mixture formed by adding 60.0 mL of 0.025 M HCl to 170 mL of 0.015 M HI. Express your answer to two decimal places.
pH = 1.75 In the mixture, both acids ionize to form H+ ions affecting the pH. Find the number of moles of each acid in the mixture by converting the volume of each solution from milliliters to liters and then using the molarity of the corresponding solution as a conversion factor between liters and moles. These conversions are summarized as follows: mol HCl = 60.0 L soln × (1 L soln / 1000 mL soln) × (0.025 mol HCl / 1 L soln) = 1.5E−3 mol HCl mol HI = 170 L soln × (1 L soln / 1000 mL soln) × (0.015 mol HI / 1 L soln) = 2.6E-3 mol HI Since both HCl and HI produce 1 mol H+ per 1 mol of ionized acid, the total number of moles of H+ in the mixture is the sum of the numbers of moles of the acids, 4.1E−3 mol. Assuming that the volume of the solution after mixing equals the sum of the volumes of the solutions before mixing, the final volume is 230 mL or 0.230 L. Now divide the number of moles of H+ by the volume of the solution in liters to find the molarity of H+ in the mixture: Molarity H+ = (mol H+ / L soln) = (4.1E−3 mol H+ / 0.230 L soln) = 1.8E−2 mol / L = 1.8E−2M Finally, substitute this concentration into the equation for pH to get pH = −log[H+] = −log(1.8E−2) = 1.75
Ammonia, NH3, is a weak base with a Kb value of 1.8×10^−5. What is the pH of a 0.250 M ammonia solution? Express your answer numerically to two decimal places.
pH = 11.33 (My answer of 11.32 was accepted) (Use Kb and ICE table to find [OH-], use that to figure out pOH and then pH) The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B B(aq)+H2O(l)⇌BH+(aq)+OH−(aq) this constant is given by Kb=[BH+][OH−] / [B] Strong bases will have a higher Kb value. Similarly, strong bases will have a higher percent ionization value.
Calculate the pH of this solution. Express your answer to two decimal places. (NOTE: [OH-] = 6.38*10^-3)
pH = 11.80 The concentration of OH− determined in Part A is used to find the pH. Since pH+pOH=14, one of the ways to calculate pH is pH = 14 + log[OH−]
Determine the pH of 0.57 M methylamine (CH3NH2) with Kb = 4.4 x 10-4 : CH3NH2(aq)+ H2O(l) ⇌ CH3NH3+ (aq) + OH- (aq).
pH = 12.2 (Use Kb to find the value for [OH-], find pOH, then use pOH to find PH)
Calculate the pH of a solution made by adding 2.70 g of lithium oxide (Li2O) to enough water to make 1.600 L of solution. Express your answer using three decimal places.
pH = 13.053 (Don't forget to take into account that Li2O results in 2 OH- forming: Li2O + H2O --> 2LiOH. LiOH is a strong base so it will completely dissociate. Calculate the number of moles of Li2O in 2.70 g then find the molarity by dividing that number by 1.6L. Multiply the molarity of Li2O by 2 to get the concentration of OH- (remember there are 2 mol of OH- produced for every 1 mol of Li2O). Use the [OH-] to find pOH (pOH = -log[OH-]) Then solve for pH using: pH + pOH = 14.00)
Determine the pH of 0.036 M formic acid (HCO2H), Ka = 1.8 x 10-4? HCO2H (aq) ⇌ H+ (aq) + HCO2- (aq) pH = 12.6 pH = 2.6 pH = 1.4 pH = 0.26
pH = 2.6 (Formic Acid is a weak acid so use an ICE table to find the equilibrium concentrations then plug those values into Ka = [H+][HCO2-] / [HCO2H])
The active ingredient in aspirin is acetylsalicylic acid (HC9H7O4), a monoprotic acid with a Ka of 3.3×10−4 at 25 ∘C. What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, containing 480 mg of acetylsalicylic acid each, in 330 mL of water? Express your answer to two decimal places.
pH = 2.67 (To solve, you are given 2 tablets with 480 mg of the acid which is 960 mg in total. Converted to grams it is 0.960g. Acetylsalicylic acid has a molar mass of 180.159 so in 0.960g there is 5.32863x10^-3 moles. The molarity will be 5.32863*10^-3 / .330L = 1.614735x10^-2 M This is the initial concentration of the acid. Plug this into an ICE table and solve for x which will be the [H+]. X will equal .0021492. Plug that value into the formula for pH: pH = -log[H+] to get the answer of 2.6677
Calculate the pH of each of the following strong acid solutions. 2.09×10^−3M HNO3 Express your answer to three decimal places.
pH = 2.680 Since HNO3 is a strong acid, it ionizes completely as per the following equation: HNO3(aq)→H+(aq)+NO3−(aq) According to this equation, one mole of the acid produces one mole of H+ ions. Therefore, the concentration of H+ ions in this solution numerically equals the concentration of HNO3. Substitute this value into the equation for pH to get pH = −log[H+] = −log(2.09×10^−3) = 2.680
Calculate the pH of each of the following strong acid solutions. 0.240 g of HClO3 in 4.00 L of solution. Express your answer to three decimal places.
pH = 3.148 Since HClO3 is a strong acid, it ionizes completely as per the following equation: HClO3(aq)→H+(aq)+ClO3−(aq) According to this equation, one mole of the acid produces one mole of H+ ions. Therefore, the concentration of H+ ions in this solution numerically equals the concentration of HClO3. To find the molarity of HClO3, begin by calculating the number of moles from the given mass using the molar mass of HClO3 (84.46 g/mol) as a conversion factor: mol HClO3 = 0.240g HClO3 × (1 mol HClO3 / 84.46 g HClO3) = 2.84E−3 mol HClO3 Then find the molarity of HClO3 by dividing the number of moles by the volume of the solution: Molarity HClO3 = mol HClO3 / L soln = 2.84E−3 mol HClO3 / 4.00 L soln = 7.10E−4 mol/L = 7.10E−4 M Since the concentration of H+ ions in this solution numerically equals the concentration of HClO3, substitute the concentration value into the equation for pH to get pH = −log[H+] = −log(7.10E−4) = 3.148
Arrange the following molecules/ions from the weakest to strongest base: Cl−, OH−, O2−, NH3. weakest Cl− < NH3 < O2− < OH− strongest weakest Cl− < NH3 < OH− < O2− strongest weakest NH3 < Cl− < OH− < O2− strongest weakest base Cl− < O2− < NH3 < OH− strongest weakest O2− < Cl− < NH3 < OH− strongest
weakest Cl− < NH3 < OH− < O2− strongest HCl is a strong acid; therefore, its conjugate base Cl− will have negligible basicity in water and will be the weakest base in the list. NH3 is a weak base and OH− is a strong base, so OH− will be the stronger of the two. Finally, O2− is the conjugate base of OH− and therefore must be the strongest base in the list.
A particular nitric acid solution has an H+ concentration of 6.7 M. What is the pH of this solution?
−0.83 [H+] = 6.7 and pH = −log(6.7) = −0.83. It may seem strange that the pH is negative, but when the concentration of a strong acid is greater than 1.0 M, the pH becomes negative.
