Chem 207 Ch.14 & 16
A diprotic acid, H₂A, has Ka1 = 3.4 × 10⁻⁴ and Ka2 = 6.7 × 10⁻⁹. What is the pH of a 0.18 M solution of H₂A?
- First, set up an ICE table. - Write the equilibrium expression and plug in values to solve for x, which is equal to the concentration of [H+][HX+].3.4×10−4=[x][x][0.18]3.4×10−4=[x][x][0.18]x=7.82304288624×10−3x=7.82304288624×10−3 - You can ignore the contribution from the second ionization because it is so much smaller than the contribution from the first ionization. - Next, find the pHpH. −log[H+]=pH−log[HX+]=pH−log[7.82304288624×10−3]=2.11
Collision Theory
- For a reaction to occur, the particles must collide, they must collide with the appropriate orientation, and they must collide with sufficient energy - Reactions will only occur if the molecules collide with energy > activation energy
Catalysis
- Increases the rate by providing a mechanism with a lower activation energy - because the activation energy is lower a greater proportion of collisions occur with energy > Ea
Lewis Acid Base Concept
- Lewis Acid: electron pair acceptor (H+ always acts as an electron pair acceptor, nonmetal oxides) - Lewis Base: electron pair donor (metal oxides)
Activation energy (Ea)
- The amount of energy required to produce the transition state of a chemical reaction. - If the activation energy for a reaction is very high, the reaction occurs very slowly - Enzymes (and other catalysts) increase reaction rates by reducing activation energy
Reaction Mechanisms
- The series of elementary reactions or steps that describe the process by which a reaction occurs at the molecular level - The mechanism must match the experimental determined rate law
What is the concentration of a reactant after 26.5 s if the initial concentration is 0.150 M and the rate constant is 5.4 x 10⁻² s⁻¹?
0.036M - The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .The units s⁻¹ indicate that this is a first-order reaction. Use the integrated rate law to solve for the concentration after 22.0 seconds.[A]=[Ao]e−kt[A]=[Ao]e−kt[A]=[0.150]e−(5.4×10−2)(22.0)[A]=[0.150]e−(5.4×10−2)(22.0)[A]=0.0457M
What is the rate for the second order reaction A → Products when [A] = 0.329 M? (k = 0.761 M⁻¹s⁻¹)
0.08237 M/s
What is the rate for the first order reaction A → Products when [A] = 0.258 M? (k = 0.360 1/min)
0.09288 M/min
The half life of the element X is 500 years. If there are initially 8g of x, how much will remain after 1500 years?
1g; because 8/2=4 and 4/2=2 and 2/2=1
Weak Acids
Acids that are only slightly ionized in aqueous solutions
Given the reaction A + 2 B → 3 C, the rate of appearance of C is also equal to which of the following?
B) -3/2∆[B]/∆t
Which one of the following 1.0 M solutions would have the highest pH? a. HF b. C₆H₅COOH c. HCNO d. HNO2
B) C₆H₅COOH - The solution with the lowest Ka value would be the weakest acid and have the highest pH out of this list. That is C₆H₅COOH
Consider the table of weak acids below. Which of the following bases would be the strongest weak base? a. CN- b. HO2- c. IO- d. BrO-
B) HO₂⁻ - A larger Ka value indicates a stronger acid. For a conjugate pair, this relationship is true: Ka×Kb=KwKa×Kb=Kw Therefore, the stronger the weak acid, the weaker the conjugate weak base. From this list, the weak acid with the smallest Ka value would yield the strongest conjugate base.
Given the following acid dissociation constants: Ka (H₃PO₄) = 7.5 × 10⁻³ Ka (NH₄⁺) = 5.6 × 10⁻¹⁰ Which statement is true about the following reaction? H₃PO₄(aq) + NH₃(aq) → NH₄⁺(aq) + H₂PO₄⁻(aq) a. This reaction is reactant-favored b. This reaction is product-favored c. The reaction enthalpy is required in order to determine
B) This reaction is product-favored.
The weak acid HY is much stronger than weak acid HX. Which one of the following statements is true? a. Y- is a stronger base than X- b. Y- is a weaker base than X-
B) Y⁻ is a weaker base than X⁻. For a conjugate pair, this relationship is true: Ka×Kb=KwKa×Kb=KwA large KaKa would result in a small KbKb, indicating a weak base. Therefore, the stronger the weak acid, the weaker the conjugate base.
Consider the proposed mechanism for the decomposition reaction 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g) in the presence of I⁻(aq). What is the rate law for this reaction? Step 1: Slow H2O2(aq) +I-(aq) -- H2O(l) + IO-(aq) Step 2: Fast H202(aq) + IO- (aq) -- H2O(l) + O2(g) + I-(aq)
C) rate = k[H₂O₂][I⁻] - Step one is bimolecular with a rate law: rate = k₁[H₂O₂][I⁻]. This is the slow step in the mechanism. Fast steps following the slow step do not enter into the rate law, so the overall rate law is rate = k[H₂O₂][I⁻].
All of the following are factors that affect the rate of a reaction except _____ a. the concentration of the reactants b. presence of a catalyst c. the magnitude of the equilibrium constant d. the temperature of the reaction
C) the magnitude of the equilibrium constant.
Amphiprotic
Can act as a proton acceptor or a proton donor
Which of the following solutions would be expected to have a pH greater than 7.00? a. NH4Br b. C₆H₅NH3Br c. Ca(N)3)2 d. C₆H₅COONa
D) C₆H₅COONa A pH greater than 7 indicates the presence of a proton acceptor rather than a proton donor. NH4+NHX4X+ and C6H5NH3+CX6HX5NHX3X+ are protic ions, with hydrogens attached to electronegative nitrogen, so they are proton donors. NO3−NOX3X− and C6H5COO−CX6HX5COOX− are possible proton acceptors, but NO3−NOX3X− is the conjugate base of a strong acid (nitric acid, pKa -1.5) whereas C6H5COO−CX6HX5COOX− is the conjugate base of a weak acid (benzoic acid, pKa 4). C6H5COO−CX6HX5COOX− is the strongest proton acceptor and raises the pH above 7.
Weak Acid ionization
HA(ag) + H2O(l) = H3O+(aq) = A-(aq) Ka= ([H3O+][A-]) / [HA] - Larger Ka's mean an increase in acidic strength
6 Common Strong Acids
HCl, HBr, HI, HNO3, H2SO4, HClO4
Consider the reaction below. Which species is(are) the Brønsted-Lowry acid(s)? HF (aq) + NH₃ (aq) ⇌ NH₄⁺ (aq) + F⁻ (aq)
HF, NH₄⁺ - Brønsted-Lowry bases accept protons and Brønsted-Lowry acids donate protons. In the forward reaction, the Brønsted-Lowry acid is HF. In the reverse reaction, the Brønsted-Lowry acid is NH4+NHX4X+ .
Consider the reaction below. Which species are conjugate acid/base pairs? HSO₃⁻ (aq) + HCN (aq) ⇌ H₂SO₃ (aq) + CN⁻ (aq)
HSO₃⁻, H₂SO₃ - The conjugate acid is the species that is formed after the base accepts a proton. In the forward reaction, HSO3−HSOX3X− is the base and H2SO3HX2SOX3 is the conjugate acid. In the reverse reaction, H2SO3HX2SOX3 is the acid, and HSO3−HSOX3X− is the conjugate base.
Conjugate Base
Ka x Kb = Kw - The particle that remains when an acid has donated a hydrogen ion
Which of the following statements is/are true? 1. The reaction of HCl with an equimolar amount of NH₃ will result in an acidic solution. 2. The reaction of NH₄Cl with an equimolar amount of KOH will result in a basic solution. 3. The reaction of HNO₃ with an equimolar amount of NaOH will result in an acidic solution.
Number 2 only
Weak Bases
Only ionize partially in aqueous solution Organic: (NH3, (CH3)2NH) Inorganic: any anion other than chloride, bromide, iodide, nitrate, perchlorate, and hydrogen sulfate (HPO4^2-, NO2^-)
If the first-order half-life of tritium (³H) is 12.26 years, what amount of time is necessary for it to lose 75% of its radioactivity?
Reframe the problem to consider the amount of radioactive material that still remains. If 75% is lost, that means that there is 25% remaining. The first half-life results in 50% remaining and the second half-life results in 25% remaining. If one half-life is 12.26 years, multiply this by 2 to find the amount of time. Answer: 2×12.26=24.52years
Strong Bases
Soluble hydroxide salts ex. LiOH, NaOH, KOH, Ba(OH)2
A reaction is determined to have the rate law rate = k[NO]²[H₂]. What is the rate-determining step in the mechanism? Step 1: 2NO = N2O2 Step 2: N2O2 + H2 -- N2O + H2O Step 3: N2O + H2 -- N2 + H2O
Step 2 - You can determine which step is rate-determining. To determine the rate law for a mechanism containing an equilibrium step, you must redefine the intermediate in terms of the equilibrium step. You cannot have any intermediates in the rate law. The rate determining step is Step 2.
Bronsted-Lowry Acid
The Bronsted-Lowry acid is a proton donor, while the Bronsted Lowry base is a proton acceptor
Inductive Effects
The acidity of H-A increases with the presence of electron-withdrawing groups in A.
In an aqueous solution at 25°C, if [H₃O⁺] = 1.7 × 10⁻⁴ M, then [OH⁻] is:
The chemical equation for the ionization water is shown. 2H2O ⇌H3O++OH−2HX2O ⇌HX3OX++OHX− The autoionization constant (KWKW) can be calculated using the relationship between the concentrations of H3O+HX3OX+ and OH−OH−.KW=[H3O+][OH−]=1.0×10−14KW=[HX3OX+][OHX−]=1.0×10−14 Plug the given values in to solve for [OH−][OHX−]. 1.0×10−14=[2.4×10−4][OH−]1.0×10−14=[2.4×10−4][OHX−][OH−]=4.2×10−11
The reaction X + X₃ → 2 X₂ is shown to have an activation energy of 45 kJ/mol while the enthalpy of reaction (∆H) is -388 kJ/mol. What is the activation energy of the reverse reaction?
The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .To calculate ΔHΔH, subtract the positive reverse EaEa from the positive forward EaEa. Plug the given numbers in to solve for EareverseEareverse. Answer: Eaforward−Eareverse=ΔHEaforward−Eareverse=∆H25−x=−38825−x=−388x=25+388=413
The pH of a basic solution is 12.87. What is [OH⁻]?
To determine the concentration of OH−OHX− in the solution with a pHpH of 9.77, we can first find the pOHpOH of the solution: pOH=14.00−pH=4.23pOH=14.00−pH=4.23 To find the concentration of OH−OHX− we can use the relationship: [OH−]=10−pOH[OH−]=10−pOH So if the pOHpOH = 4.23, then [OH−]=10−4.23=5.9×10−5M
What mass of KOH would need to be dissolved in 500.0 mL of water to produce a solution with a pH of 12.40?
To make 500.0 mL of a basic solution with a pHpH of 12.40 using KOHKOH, we need to determine how many moles of hydroxide will be required for the solution. We can first use pHpH to find the pOHpOH: pOH=14.00−pH=14.00−12.40=1.60pOH=14.00−pH=14.00−12.40=1.60 Now that we know the pOHpOH, we can use that to find the concentration of OH−OHX−.[OH−]=10−pOH=10−1.60=0.0251M[OH−]=10−pOH=10−1.60=0.0251M In 500.0 mL of solution we would need with (0.5000 L)(0.0251 M) = 0.0126 moles OH−OHX−. Since the source of our hydroxide ion is KOHKOH, which produces one mole of OH−OHX− per mole of KOHKOH, we then use the molar mass of KOHKOH to find the mass: (0.0126mol KOH)(56.11g/mol)=0.70 g KOH
The isomerization of cyclopropane to propene occurs with a first-order rate constant of 2.42x10^-2/hr. How long will it take for the concentration of cyclopropane to decrease from an initial concentration of 0.080mol/L to 0.020mol/L?
Using the equation, ln([R]t / [R]0)= -kt we find that 57.3hr = t
What is the activation energy for the isomerization of methyl isocyanide? Rate Constants and Temperatures 4.30 x 10⁻³ min⁻¹ 472 K 5.54 x 10⁻² min⁻¹ 503 K
You can solve for the EAEA by rearranging this equation and plugging in given values. Answer: ln(k1k2)=EAR(1T2−1T1)ln(k1k2)=EAR(1T2−1T1)EA=(R)(T1)(T2)(T1−T2)ln(k1k2)EA=(R)(T1)(T2)(T1−T2)ln(k1k2)EA=(8.314)(472)(503)(31)ln(5.54×10−24.30×10−3)EA=(8.314)(472)(503)(31)ln(5.54×10−24.30×10−3)EA=162,747J/molEA=162,747J/mol Convert this to kJ by dividing by 1000. 162,747J1000=163kJ
Integrated Rate Law
a relationship between the concentrations of the reactants in a chemical reaction and time
Will the following aqueous solutions be acidic, basic, or neutral? a. 0.10 M NaCH3COO b. 0.10 M NaBr c. 0.10 M NaHSO4 d. 0.10 M NH4NO3 e. 0.10 M K2HPO4
a. basic b. neutral c. acid d. acid e. acid AND base
Polyprotic Acid
an acid that contains more than one ionizable proton and releases them sequentially Ex. H3PO4(aq) +H2O(l) = H3O^+(aq) + H2PO4^-(aq)
Bimolecular
an elementary step in a reaction that involves two particles, either the same species or different, that collide and go on to form products
Rate Law
an expression relating the rate of a reaction to the concentration of the reactants
Which of the following is an amphoteric species? a. H2SO3 b. HSO3- c. SO3^2-
b. HSO3- - An amphoteric species is able to both donate and accept a proton.
If a plot of 1/[A] versus time produces a straight line with a positive slope for the reaction A → B + C, what is the order of the reaction? a. zero b. first c. second
c. second; A graph of 1/[A] versus time yields a straight line with a positive slope for second order reactions.
Arrhenius Equation
lnk=(-Ea/R)(1/T)+lnA - Increasing the temperature, increases the proportion of collisions that have energy > Ea
pH formula
pH= -log[H3O+]
pOH formula
pOH=-log[OH-]
Rate of reaction
the change in concentration of a reactant or product per unit time
Molecularity
the number of molecules that participate as reactants in an elementary reaction
Rate determining step
the slowest step in a reaction mechanism
Autoionization of water
when pure water reacts with itself to for hydronium and hydroxide ions -logKw=
