Chapter 16 & 17 Chemistry
ion-product constant for water (Kw)
1.0 x 10^-14 @ 25 degrees Celsius (equilibrium constant for the auto ionization of water)
Conjugate Acids and Bases
A base accepts a proton and becomes a conjugate acid An acid donates a proton and becomes a conjugate base
A small K (K << 0) would indicate:
A reaction with a lot of reactants relative to products at equilibrium (Forward reaction is not favored in such a case)
Which of the following is NOT a Brønsted-Lowry acid? Hint for Question 3 A Brønsted-Lowry acid is a proton donor.
CO3^2- Correct, carbonate ion does not contain hydrogen and cannot be a proton donor
carboxylic acid
COOH contains this group (acetic, citric, malic)
Identify the conjugate base in the reaction shown here. HClO2 (aq) + H2O (l) ⇌ H3O + (aq) + ClO −2 (aq)
ClO2-
Which of the following illustrates a conjugate acid-base pair?
HNO2 / NO2^-
The equilibrium constant for the reaction shown here is Kc = 1.0 x 10^3. A reaction mixture at equilibrium contains [A] = 1.0 x 10^-3 M. What is the concentration of B in the mixture? A (g) -> <- B (g)
Kc = [B] / [A] 1.0 x 10^3 = [B] / 1.0 x 10^-3 (1.0 x 10^3) x (1.0 x 10^-3) = [B] [B] = 1.0 M
What is the correct expression for the equilibrium constant (Kc) for the reaction between carbon and hydrogen gas to form methane shown here? C (s) + 2 H2 (g) -> <- CH4 (g)
Kc = [CH4] / [H2]^2
The reaction shown here as a KP = 4.5 x 10^2 at 825 K. Find Kc for the reaction at this temperature. CH4 (g) + CO2 (g) -> <- 2 CO (g) + 2 H2 (g)
Kp = Kc(RT)^delta n Delta n = 4 -2 = 2 4.5 x 10^2 = Kc (0.08206 L(atm)/mol(K) x 825 K)^2 Kc = 4.5 x 10^2 / (0.08206 L(atm)/mol(K) x 825 K)^2 Kc = 0.098
Which pair is a Brønsted-Lowry conjugate acid-base pair?
NH3; NH4+ The acid and conjugate base differ by H+. NH3 + H+ ⇌ NH4+
Which compound will form an acidic solution when dissolved in water?
NH4Cl NH4Cl is formed from the conjugate acid of a weak base (NH3) and the conjugate base of a strong acid (Cl−), so the solution is acidic.
Rank the following in order of increasing base strength (weakest to strongest). OI− OCl− OBr−
OCl- < OBr- < OI- Cl is more electronegative than Br, Br is more electronegative than I
What is the pH of 0.0200 M HCl? Hint: A strong acid ionizes completely; therefore, the concentration of the strong acid is equal to [H3O+].
[HCl] = 0.0200 M = [H3O+] PH = -log[H3O+] = -log[0.0200] = 1.70
triprotic acid
an acid able to donate three protons per molecule
Monophonic acid
an acid that can donate only one proton (hydrogen ion) per molecule
diprotic acid
an acid that can donate two protons per molecule [X^2-] = Ka2
An HNO3(aq) solution has a pH of 1.75. What is the molar concentration of the HNO3(aq) solution?
pH - -log[H3O+] [H3O+] = 10^-pH = 10 ^ -1.75 = 0.018 M
autoionization
the process by which water acts as an acid and a base with itself
Amphotenc
Ability to act as an acid or a base
Which of the following statements about the equilibrium constant is false? A.) When adding chemical equations, multiply the equilibrium constants by each other to determine the overall equilibrium constant. B.) If the coefficients in an equation are multiplied by three, the equilibrium constant should also be multiplied by 3 C.) The equilibrium constant of the reverse reaction is equal to the reciprocal of the equilibrium constant for the forward direction D.) Doubling the amount of a solid in a reaction never has an effect on the equilibrium constant
B.) If the coefficients in an equation are multiplied by three, the equilibrium constant should also be multiplied by 3 (The equilibrium constant should be raised to the third power (cubed))
For the reaction 2 A (g) -> <- B (g), the equilibrium constant for Kp = 0.76. A reaction mixture initially contains 2.0 atm of each gas (PA = 2.0 atm and PB = 2.0 atm). Which statement is true of the reaction mixture? A.) The reaction mixture will proceed toward the reactants B.) The reaction mixture will proceed toward products C.) The reaction mixture is at equilibrium D.) It is not possible to determine the future direction of the reaction mixture from the information given.
B.) The reaction mixture will proceed toward the products Qp = PB / PA = 2.0 / (2.0)^2 = 0.50 0.50 < 0.76 Qp is less than Kp therefore the reaction mixture proceeds toward the products
What is the [OH−] of a 0.200 M solution of ethylamine (C2H5NH2)? For ethylamine, Kb = 5.6 × 10−4.
C2H5NH2- (aq) + H2O (l) -> <- OH^- (aq) + C2H5NH3 + (aq) Initial: [C2H5NH2]: 0.200, [OH-]: 0.000, [C2H5NH3+]: 0.000 Change: [C2H5NH2]: -x, [OH-]: +x, [C2H5NH3+]: +x Equilibrium: [C2H5NH2]: 0.200 - x, [OH-]: x, [C2H5NH3+]: x Kb = x^2/0.200 - x 5.6 x 10^-4 = x^2 / 0.200 - x Assume x << 0.200, x^2/0.200 = 5.6 x 10^-4 X = 0.011 = [OH-] % hydrolized = (0.011 M/0.200 M) x 100% = 5.3% so assumption is good.
Calculate the pH of a 0.500 M pyridine (C5H5N) solution. Kb value for pyridine is 1.7 × 10−9.
C5H5N (aq) + H2O (l) -> <- CH5NH+ (aq) + OH- (aq) I: 0.500 M, 0, 0 C: -x, +x, +x E: 0.500-x, x, x Kb = 1.7 x 10^-9 1.7 x 10^-9 = x^2/0.500-x Assume x << 0.500 1.7 x 10^-9 = x^2/0.500 X = 2.9 x 10^-5 = [OH-] pOH = -log[OH-] = -log[2.9 x 10^-5] = 4.54 pH = 14.00 - 4.54 = 9.46
Which of the following is true about the role of solids and liquid in equilibrium processes? A.) The concentration of a solid increases as a solid is precipitated out of solution. B.) The concentration of a pure liquid must be included in the equilibrium constant expression. C.) The concentrations of solids and pure liquids will increase with an increase in temperature. D.) The concentration of a solid or a pure liquid is constant as long as some of the pure substance is present. E.) The concentration of a pure liquid will change with the addition of a solute.
D.) The concentration of a solid or a pure liquid is constant as long as some of the pure substance is present. (Concentration depends only on density)
Consider the reaction of A to form B. 2 A (g) -> <- B (g) Kc = 1.8 x 10^-5 (at 298 K) A reaction mixture at 298 K initially contains [A] = 0.50 M. What is the concentration of B when the reaction mixture reaches equilibrium?
Initial: [A]: 0.50 M, [B] = 0.00 M Change: [A] = 0.50 - 2x, [B] = +x Equilibrium: [A] = 0.50 - 2x, [B] = x Kc = [B] / [A]^2 1.8 x 10^-5 = x / (0.50 - 2x)^2 (1.8 x 10^-5) times (0.50 -2x)^2 = x Assume x is <<< 0.50 X = 4.5 x 10^-6 is less than 0.50 therefore it is good
At 445 ºC calculate the equilibrium concentration of HI when 0.500 M HI, 0.0200 M H2, and 0.0200 M I2 are placed in a sealed flask and allowed to reach equilibrium. Kc for this reaction is 50 at 445 ºC. H2 (g) + I2 (g) ⇆ 2 HI (g)
Qc = (0.500)^2 / (0.0200)(0.0200) = 625 > Kc shift toward the reactant meaning product decreases Initial: [H2]: 0.0200, [I2]: 0.0200, [HI]: 0.500 Change: [H2]: +x, [I2]: +x, [HI]: -2x Equilibrium: [H2]: 0.0200 + x, [I2]: 0.0200 + x, [HI]: 0.500 -2x 50 = (0.500 - 2x)^2/(0.0200 + x)^2 Square root 50 = 0.500 -2x / 0.0200 + x 7.07 (0.0200+ x) = 0.500 - 2x 0.141 + 7.07x = 0.500 - 2x -0.359 = -9.07x X = 0.0396 0.500 - 2(0.0396) = 0.4208 = 0.421 M
In which direction will the following reaction proceed when PCO = 0.13 atm, PH2O = 0.56 atm, PCO2 = 0.62 atm, and PH2 = 0.43 atm. Kp for the reaction is 2.7. CO (g) + H2O (g) ⇆ CO2 (g) + H2 (g)
Qp = (0.62) (0.43) / (0.13) (0.56) Qp = 3.7 > Kp Toward the reactants
pKa = -logKa
The smaller the pKa, the stronger the acid
PKb = -logKb
The smaller the value of Kb, the weaker the base
Strong acids mean
The stronger the acid the weaker the conjugate base
Determine the pH of 0.450 M HC7H5O2. The Ka for HC7H5O2 is 6.5 × 10−5.
Use Ka Equilibrium: HA -> <- H+ + A- Ice table calculation I: [HA]: 0.450, [H+]: 0, [A-]: 0 C: -x, +x, +x E: 0.450 -x, x,x Ka = x^2/0.450 -x 6.5 x 10^-5 = x^2/0.450-x Use quadratic equation x = 0.00537592 PH = -log(5.4 x 10^-3) = 2.267 = 2.27
Percent ionization
% = concentration ionized acid / initial concentration of acid x 100%
Bases
- Bittery taste - Slippery feels - Red litmus paper blue - Neautralize acids
Which of the following is considered an Arrhenius acid? Hint for Question 2 An Arrhenius acid is defined as a substance that produces H+ ions in aqueous solution.
H2SO4 (Correct, sulfuric acid is also a Brønsted-Lowry acid.) According to Bronsted-Lowry theory, an acid is proton donor and base is proton acceptor. According to Arrhenius theory, hydrochloric acid is an acid which gives hydrogen ions in water but according to Bronsted-Lowry theory, hydrochloric acid is an acid because it donates a proton to the water molecule
The reaction X2 (g) -> <- 2X (g) occurs in a closed reaction vessel at constant volume and temperature. Initially, the vessel contains only X2 at a pressure of 1.55 atm. After the reaction reaches equilibrium, the total pressure is 2.85 atm. What is the value of the equilibrium constant, Kp, for the reaction?
Initial: [PX2] = 1.55 atm, [PX] = 0.00 atm Change: [PX2] = -x. [PX] = +2x Equilibrium: [PX2] = 1.55 - x, [PX] = 2x Total pressure: (1.55 - x) + 2x = 2.85 Total pressure: x = 1.30 Kp = [PX]^2 / [PX2] Kp = (2x)^2 / 1.55 -x Kp = (2 x 1.30)^2 / (1.55 - 1.30) Kp = 27
Changes in equilibrium constant
Reverse: Invert (1/Kc) Multiply: Raise to that factor (K^n) Adding 2 or more together: Koverall = [P]/ [R]
What is the pH of a solution of 0.0350 M Ba(OH)2?
[Ba(OH)2] = 2[OH-] = 0.07 M POH = -log(0.07) = 1.15 PH = 14.00 - 1.15 = 12.85
The Ka value for a weak acid is _________ than the Ka value for a strong acid because the weak acid ionizes to a _________ extent than the strong acid.
smaller; lesser Correct, there are fewer products at equilibrium in the ionization of a weak acid.
Which of the following will NOT cause the reaction to shift toward the products? H2 (g) + I2 (g) ⇆ 2 HI (g) ΔH = -9.42 kJ
Decreasing the volume of the container (changing the volume of the container does not affect this reaction since the number of moles of gaseous product is the same as the number of moles of gaseous reactants.)
Write the equilibrium constant expression for the following reaction: CH4 (g) + 2 H2S (g) -> <- CS2 (g) + 4 H2 (g)
K = [CS2] [H2]^4 / [CH4] [H2S]^2
What is the concentration of X^2− in a 0.150M solution of the diprotic acid H2X, Ka1 = 4.5 × 10^-6 and Ka2 = 1.2 x 10^-11.
Ka2 = 1.2 x 10^-11 << Ka1 = 4.5 x 10^-6, [X^2-] = Ka2 = 1.2 x 10^-11 M
What is the pH of a 0.250 M NH4Cl solution? Kb for NH3 = 1.8 × 10−5.
NH4Cl -> NH4+ + Cl- NH4+ (aq) + H2O (l) -> <- H3O+ (aq) + NH3 (aq) pH = -log[H3O+] Ka = Kw/ Kb = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10 NH4+, H3O+, NH3 I: 0.250, 0, 0 C: -x, +x, +x E: 0.250 -x, x, x Ka = x^2 / 0.250 -x =5.56 x 10^-10 Assume x << 0.250 X^2/0.250 = 5.56 x 10^-10 X = 1.18 x 10^-5 = [H3O+] pH = -log[H3O+] = -log[1.18 x 10^-5] = 4.93
Calculate the reaction quotient for the reaction below where the concentrations are [SO3] = 0.500 M, [SO2] = 0.0500 M, and [O2] = 0.100 M. 2 SO3 (g) ⇆ 2 SO2 (g) + O2 (g)
Qc = (0.0500)^2 (0.100) / (0.500)^2 Qc = 1.0 x 10^-3
Use the data below to find the equilibrium constant (Kc) for the reaction A (g) -> <- 2 B (g) + C (g): A (g) -> <- 2 X (g) + C (g) Kc = 1.55 B (g) -> <- X (g) Kc = 25.2
Reverse the second reaction and double it 2 X (g) -> <- 2 B (g) To the second reaction's Kc, square it and inverse it (1/25.2)^2 = 0.001574704 1.55 times (1/25/2)^2 = 2.44 x 10^-3
Which ion will be basic in aqueous solution?
SO3^2- SO3^2- is the conjugate base of a weak acid, so it is basic in an aqueous solution
[H3O+] in a strong acid solution is equal to the concentration of the strong acid
STRONG ACIDS COMPLETELY IONIZE IN WATER
Strong base completely dissolves in solution
Sr(OH)2 (aq) -> <- Sr 2+ (aq) + 2 OH- (aq)
Acids
- Have sour taste - The ability to dissolve many metals - Turn blue litmus paper red - Neutralize bases
The following reaction has a Kc value of 4.1 × 10-31 at 25 ºC. Calculate Kp at this temperature. N2 (g) + O2 (g) ⇆ 2 NO (g)
4.1 x 10^-31 Because delta n equals 0
Which of the following is false about acids and bases? All acids have a pH lower than 7. A weak base in equilibrium with water will produce the hydronium ion. A base can be defined as a substance that accepts protons. Relative acid strength can be determined based on Ka values. Strong acids and bases completely ionize in water.
A weak base in equilibrium with water will produce the hydronium ion. (This is the effect of a weak acid)
Consider the given acid ionization constants and identify the strongest conjugate base. Acids and Ka: HNO2 (aq); 4.6 x 10^-4 HCHO2 (aq); 1.8 x 10^-4 HClO (aq); 2.9 x 10^-8 HCN (aq); 4.9 x 10^-10
CN- (aq) The weaker the acid (the smaller the Ka), the stronger the conjugate base. HCN has the smallest Ka, so it has the strongest conjugate base (CN−).
If Kc for the following reaction is 5.4 x 10^13, 2 NO (g) + O2 (g) -> <- 2 NO2 (g) What is the Kc for the reaction shown below? 4 NO2 (g) -> <- 4 NO (g) + 2 O2 (g)
Double the first reaction and reverse it This means the the Kc become the reciprocal and squared: (1 / (5.4 x 10^13)^2)= 3.4 x 10^-28
Predict the products of the reaction below: HNO2 + H2O ⇆
NO2^- + H3O^+ Water acts as the base in this reaction
Which of the following salts would produce a basic solution? KI NaC2H3O2 NH4Cl Ca(NO3)2
NaC2H3O2 The acetate anion acts as a base
Neutral, Acidic, or Basic?
Neutral- H3O+ and OH- are equal pH = 7 Acidic - creates additional H3O+ ions causing [H3O+] to increase > [OH-] pH < 7 Basic - creates additional OH- ions causing [OH-] to increase > [H3O+] pH > 7
A container is charged with 2.00 mol of I2 and is allowed to reach equilibrium. The I2 decomposes to iodine atoms as shown in the following reaction: I2 (g) ⇆ 2 I (g) After the reaction reaches equilibrium, an inert gas is introduced into the container under conditions of constant volume. Which of the following would you expect to happen?
No change; the reaction is at equilibrium
The solid XY decomposes into gaseous X and Y: XY (s) -> <- X (g) + Y (g) Kp = 4.1 (at 0 degrees Celsius) If the reaction carried out in a 22.4-L container, which initial amounts of X and Y result in the formation of solid XY?
None of the above At STP, 1 mol of gas occupies 22.4 L. In order for Qp > Kp, the product of the number of mol of X and the number of mol of Y must exceed 4.1. None of the answer options meet this criterion.
Equilibrium [H3O+] concentration of a weak acid increases with increasing initial concentration
Percent ionization of a weak acid decreases with increasing concentration HA (aq) + H2O (l) -> <- H3O+ (aq) + A- (aq) (WEAK ACID) Ka strength depends on this equation
Direction of reaction Q VS K:
Q < K = reaction goes to right (products) Q = K Reaction is at equilibrium Q > K Reaction goes to left (reactants
Which of the illustrations represents the weakest acid?
Leftmost figure because fewer molecules are ionized
Which of the following is false? Increasing the temperature causes an exothermic reaction to produce more reactants. Decreasing the temperature causes an exothermic reaction to produce more products. Increasing the temperature of an endothermic reaction increases the equilibrium constant. All of the above statements are true. None of the above statements are true.
All of the above statements are true because they correctly describe the effect of temperature
Calculate Kc for the following reaction at 300 K where the equilibrium concentrations are [SO3] = 0.400 M, [SO2] = 0.200 M, and [O2] = 0.600 M. 2 SO3 (g) ⇆ 2 SO2 (g) + O2 (g)
Kc = (0.200)^2 (0.600) / (0.400) Kc = 0.15
What is the OH− concentration in an aqueous solution at 25℃ in which [H3O+] = 1.9 × 10−9 M?
Kw = [H30+][OH-] = 1.0 X 10^-14 [OH-] = 1.0 x 10^-14 / 1.9 x 10^-9 = 5.3 x 10^-6 M
Consider two aqueous solutions of nitrous acid (HNO2). Solution A has a concentration of [HNO2] = 0.55 M, and solution B has a concentration of [HNO2] = 1.25 M. Which statement about the two solutions is true?
Solution A has the higher percent ionization and the higher pH As the concentration of a weak acid decreases, the percent ionization increases (the assumption that x is small is less likely to be valid). The lower the concentration of an acid (either weak or strong), the higher the pH.
Calculating pH
Strong acid solutions -use: pH = - log[acid] Strong base solutions -use: pH = 14.00+ log[base] Pure weak acid sol'ns -use: Ka equilibrium, ICE calculation of [H+] 4) Pure weak base sol'ns -use: Kb equilibrium, ICE calculation of [OH-] Buffer solutions: -use Ka equil. ICE calc'n of [H+] OR, use buffer equation: pH = pKa + log {[base]/[acid]}
What is the effect of adding helium gas (at constant volume) to an equilibrium mixture of this reaction: CO (g) + Cl (g) -> <- COCl2 (g)
The reaction does not shift in either direction Helium is not part of this equilibrium reaction, so adding helium (an inert gas) has no effect on the reaction
As shown by the equations: H2O + HCO3− ⇆ H3O+ + CO32− H2O + HCO3− ⇆ H2CO3 + OH−
Water is amphoteric because it can act as an acid or a base
Which of the following is a false statement about the relationship between Q and K?
When Q = K, the reaction has stopped (It just means the system is at equilibrium)
The decomposition of NH4HS is endothermic: NH4HS (s) -> <- NH3 (g) + H2S (g) Which change to an equilibrium mixture of this reaction results in the formation of more H2S? A.) a decrease in the volume of the reaction vessel (at constant temperature) B.) an increase in temperature C.) an increase in the amount of NH4HS in the reaction vessel D.) all of the above
B.) an increase in temperature Increasing the temperature of an endothermic reaction shifts the reaction to generate more products, forming more H2S.
Find the pH of 0.175 M NaCN solution. For HCN, Ka = 4.9 × 10−10.
CN- (aq) + H2O (l) -> <- OH^- (aq) + HCN (aq) Initial: [CN-]: 0.175, [OH-]: 0.000, [HCN]: 0.000 Change: [CN-]: -x, [OH-]: +x, [HCN]: +x Equilibrium: [CN-]: 0.175 - x, [OH-]: x, [HCN]: x Kb = Kw/Ka = 1.0 x 10^-14 / 4.9 x 10^-10 = 2.0408 x 10^-5 x^2 / 0.175 - x = 2.0408 x 10^-5 Assume x << 0.175 X^2/0.175 = 2.0408 x 10^-5 x = 0.0018898 = [OH-] % hydrolized = (0.0018898 M / 0.175 M) x 100% = 1.1% so assumption is good pOH = -log(0.0018898) = 2.72 pH+pOH = 14.00 pH = 14.00 - 2.72 = 11.28
What is meant by the term "dynamic equilibrium"? A.) when the concentrations of reactants and products are equal to one another B.) when the reaction equation is stiochiometrically balanced C.) when the reactants have just begun to form products D.) when the rates of the forward and reverse reactions are the same E.) when the forward and reverse reactions stop
D.) when the rates of the forward and reverse reactions are the same (No net change in the system)
Find the pH of a 0.350 M aqueous benzoic acid solution. For benzoic acid, Ka = 6.5 × 10−5.
HA (aq) + H2O (l) -> <- H3O+ (aq) + A- (aq) Initial: [HA]: 0.350, [H3O+]: 0.000, [A-]: 0.000 Change: [HA]: -x, [H3O+]: +x, [A-]: +x Equilibrium: [HA]: 0.350 - x, [H3O+]: x, [A-]: x 6.5 x 10^-5 = x^2 / 0.350 - x Assume x << 0.350 6.5 x 10^-5 = x^2 / 0.350 X = 4.8 x 10^-3 M = [H3O+] % ionization = (4.8 x 10^-3 M / 0.350 M) x 100% = 1.4% so assumption is valid pH = -log[H3O+] = -log(4.8 x 10^-3) = 2.32
What is the percent ionization of 0.0550 M solution of HC2H3O2? Hint: Percent ionization is the ratio of concentration of the ionized acid to the initial concentration of the acid expressed as a percent.
HC2H3O2 (aq) + H2O (l) -> <- H3O+ (aq) + C2H3O2^- (aq) Initial: [HC2H3O2]: 0.0550, [H3O+]: 0.000, [C2H3O2^-]: 0.000 Change: -x , +x, +x Equilibrium: 0.0550 -x, x, x 1.8 x 10^-5 = x^2 / 0.0500 -x Assume x << 0.0500, x^2/1.45 = 1.8 x 10^-5 X = 9.9 x 10^-4 M = [H3O+] % ionization = (9.9 x 10^-4 M / 0.0550 M) x 100% = 1.8% answer is 1.7%
Calculate the percent ionization of 1.45 M aqueous acetic acid solution. For acetic acid, Ka = 1.8 × 10−5.
HC2H3O2 (aq) + H2O (l) -> <- H3O+ (aq) + C2H3O2^- (aq) Initial: [HC2H3O2]: 1.45, [H3O+]: 0.000, [C2H3O2^-]: 0.000 Change: -x , +x, +x Equilibrium: 1.45 -x, x, x 1.8 x 10^-5 = x^2 / 1.45 -x Assume x << 1.45, x^2/1.45 = 1.8 x 10^-5 X = 5.1 x 10^-3 M = [H3O+] % ionization = (5.1 x 10^-5 M / 1.45 M) x 100% = 0.35%
Which acid has the largest Ka: HClO2(aq), HBrO2(aq), HIO2(aq)?
HClO2 (aq) For oxyacids, acid strength increases with increasing electronegativity of the atom bonded to the oxygen atoms. The electronegativity trend is Cl > Br > I, so chlorous acid is the strongest acid (largest Ka).
Find the pH of a 0.155 M HClO2(aq) solution. For HClO2; Ka = 0.011.
HClO2 (aq) + H2O (l) -> <- H3O+ (aq) + ClO2- (aq) Initial: [HClO2]: 0.155, [H3O+]: 0.000, [ClO2-]: 0.000 Change: [HClO2]: -x, [H3O+]: +x, [ClO2-]: +x Equilibrium: [HClO2]: 0.155-x, [H3O+]: x, [ClO2-]: x 0.011 = x^2 / (0.155-x) X^2 = 0.011x - 0.001705 X^2 + 0.011x - 0.001705 = 0 Use quadratic formula x = -b +- square root b^2 - 4ac / 2a X=0.036 M = [H3O+] pH = -log[H3O+] = -log(0.036) = 1.44
Consider the reaction between NO and Cl2 to form NOCl. 2 NO (g) + Cl2 (g) -> <- 2 NOCl (g) A reaction mixture at a certain temperature initially contains only [NO] = 0.50 M and [Cl2] = 0.50 M. After the reaction comes to equilibrium, the concentration of NOCl is 0.30 M. Find the value of the equilibrium constant (Kc) at this temperature.
Initial: [NO] = 0.50 M, [Cl2] = 0.50 M, [NOCl] = 0.00 M Change: [NO] = -2x = -(2 x 0.15) = - 0.30 , [Cl2] = -x = -0.15, [NOCl] = + 2x Equilibrium: [NO] = 0.50 M - 0.30 M = 0.20 M, [Cl2] = 0.50 M - 0.15 M = 0.35 M, [NOCl] = 0.30 M X = 0.30 M / 2 = 0.15 M Kc = [NOCl]^2 / [NO]^2 [Cl2] Kc = (0.30)^2 / (0.20)^2 (0.35) Kc = 6.4
Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. I2 (g) + Cl2 (g) -> <- 2 ICl (g) Kp = 81.9 (at 298 K) A reaction mixture at 298 K initially contains PI2 = 0.25 atm and PCl2 = 0.25 atm. What is the partial pressure of iodine mono chloride when the reaction reaches equilibrium?
Initial: [PI2] = 0.25 atm, [PCl2] = 0.25 atm, [PICl] = 0.00 atm Change: [PI2] = -x, [PCl2] = -x, [PICl] = +2x Equilibrium: [PI2] = 0.25 - x, [PCl2] = 0.25 - x, [PICl] = 2x Kp = [ICl]^2 / [I2][Cl2] 81.9 = (2x)^2 / (0.25 - x )^2 Square root 81.9 = 2x / 0.25 - x 9.04986 (0.25 - x) = 2x 2.26225 - 9.04986x = 2x 2.26225 = 11.04986 x X = 0.20475 PICl = 2x = 2(0.20475) = 0.41 atm
Forward or reverse action favored
K << 1 = small equilibrium, Num < denom. REVERSE REACTION FAVORED. FOWARD reaction does not proceed far K = 1. NEITHER is favored. Forward reaction proceeds about halfway K >> 1 = large equilibrium, Num > Denom. FOWARD REACTION favored. Proceeds to completion
Which of the following statements about the ionization of a polyprotic acid, such as H3PO4, is false? Ka1 is much larger than Ka2 and Ka3<.sub>. Ka1 = Kw We can assume [H2PO4−] = [H3O+]. [HPO42−] ≈ Ka2 The protons are removed from the polyprotic acid in a stepwise fashion.
Ka1 = Kw The magnitude of Ka1 will vary within the strength of the acid
