CHEMISTRY 1AA3 PROBLEM SET 2 – WEEK OF JANUARY 21, 2002 SOLUTIONS

1. What is the conjugate acid of each of the following species? Base Conjugate acid

(a) S2− HS− (b) CH3COO− CH3COOH (c) ClO2

− HClO2 (d) SO3

2− HSO3−

2. What is the conjugate base of each of the following species? Acid Conjugate base

(a) H2CO3 HCO3−

(b) HBr Br− (c) H2PO4

− HPO42−

(d) [(CH3)2NH2]+ (dimethylammonium ion) (CH3)2NH2 (dimethylamine)

3. Write balanced chemical reactions and Ka or Kb expressions for the first ionization of the following acids and bases (i.e. assume in each case that only one H+ is ionized or added).

(a) HF (b) H3PO4 (c) CH3NH2 (d) CH3COOH (e) HClO3

(a) HF + H2O F− + H3O+ Ka = [F−][ H3O+] [HF]

(b) H3PO4 + H2O H2PO4

− + H3O+ Ka = [H2PO4−][ H3O+]

[H3PO4]

(c) CH3NH2 + H2O CH3NH3

+ + OH− Kb = [CH3NH3+] [OH−]

[CH3NH2] (d) CH3COOH + H2O H3O+ + CH3COO− Ka = [H3O+] [CH3COO−] [CH3COOH]

(e) HClO3 is a strong acid and is fully dissociated at all concentrations in water. There is no equilibrium expression to describe the behaviour of this acid.

2 4. (a) What is the pH of a 0.035 M solution of Ba(OH)2 (aq)?

(b) What is the pH of a 0.0042 M solution of HClO4 (aq)? (c) What is the pOH of the solution in part (b)? (d) A 30.0 mL sample of HClO4 was diluted to 500 mL, and the pH of the final

solution was 4.56. What was the concentration of HClO4 in the original sample? SOLUTION:

(a) Ba(OH)2 (aq) is a soluble hydroxide, and a strong base. It dissociates fully in water, so Ba(OH)2 (aq) produces Ba2+ (aq) and 2 −OH (aq).

The original 0.035 M sample of Ba(OH)2(aq) therefore produces

0.035 M Ba2+ ions, and 2 × (0.035M) −OH ions (according to the formula of our hydroxide).

∴ [−OH] = 2(0.035M) = 0.070 M From this we can find pOH: pOH = -log10[−OH] = -log10[0.070] = 1.15 From pH + pOH = 14, rearrange to get pH = 14 – pOH = 14 – 1.15 = 12.85 = 12.9

(b) HClO4 is a strong acid, and dissociates fully in water to give:

HClO4(aq) + H2O(l) → ClO4−(aq) + H3O+(aq)

The original 0.0042 M sample of HClO4(aq) therefore produces

0.0042 M ClO4− ions, and

0.0042 M H3O+ ions pH = -log10[H3O+] = -log10(0.0042) = 2.38 (c) pOH = 14 – pOH = 14 – 2.38 = 11.62 = 11.6 (d) Since HClO4 is a strong acid, it dissociates fully in solution. In the final solution, pH = 4.56, and we can find [H3O+] directly from pH:

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Since pH = −log10[H3O+], rearrange to give −pH = log10[H3O+] and take the antilog to give 10−pH = [H3O+] Thus, [H3O+] = 10−4.56 = 2.75 × 10−5 mol/L

To find the initial concentration of HClO4, we need the number of moles of HClO4 present in solution:

mol HClO4 = (2.75 × 10−5 mol/L)(0.500 L) = 1.38 × 10−5 mol Original concentration of HClO4 = mol HClO4 original volume

= 1.38 × 10−5 mol = 4.60 × 10−4 M 0.0300 L 5. (a) What is the pH of a 10.0 M solution of HCl?

(a) What is the pH of a 1.00 × 10−8 M solution of HCl? SOLUTION:

(a) pH = −log10[H3O+] = −log10(10.0) = −1 Very acidic! The point of this question is just to show that the pH scale doesn't just go from 0 to 14, it's just that most of the substances we talk about lie within that range.

(b) Again, pH = −log10[H3O+]

BUT in this case, what is [H3O+]? Remember that from the autoionization of water, there is a concentration of

[H3O+] = 10−7 M (which we usually ignore in our calculations). In this case, however, the concentration is very significant. Thus, we have two sources of H3O+ in our solution: the H2O and the HCl: ∴[H3O+] = 1.00 ×10−8 M (from HCl) + 1.00 × 10−7 M (from water) = 1.10 × 10−7 M

Now we can use pH = −log10[H3O+] = −log10(1.10 × 10−7) = 6.96

(If you ignore the [H3O+] from water in this case, the pH will work out to be 8, which doesn't make sense, since we're talking about a dilute solution of acid!).

4 6. Dimethylamine, (CH3)2NH, is a weak base (ionization constant Kb = 7.40 × 10−4).

(a) What is the equilibrium concentration of dimethylammonium ion, [(CH3)2NH2]+ in a 0.400 M aqueous solution of (CH3)2NH?

(b) What is the pH of a 0.400 M aqueous solution of (CH3)2NH?

SOLUTION: (a) This is a “weak base” problem. Start by writing down the chemistry: (CH3)2NH + H2O (CH3)2NH2

+ + −OH Set up an “ICE” table (Initial, Change, Equilibrium concentrations) in order to obtain mathematical expressions that describe the reaction. (CH3)2NH (CH3)2NH2

+ −OH Initial conc. 0.400 M 0 (10−7 M from water)* Change -x +x +x Equilibrium conc. 0.400-x x x * We ignore the amount of −OH present from the ionization of water, if the value (10−7 M) is very small, as compared to x. Since (CH3)2NH functions as a base in our equilibrium, set up a Kb expression:

Kb = [(CH3)2NH2+][ −OH] = 7.4 × 10−4

[(CH3)2NH]

Substitute the equilibrium concentration values from the “ICE” table into Kb:

(x)(x) = 7.4 × 10−4 (0.40-x)

Assume x << 0.400, and thus (0.400-x) ≈ 0.400

x2 = (7.4 × 10−4)(0.400) x = 0.0172 M

Check: 0.0172 × 100% = 4.30% <5%, assumption is okay 0.400

∴ Equilibrium [(CH3)2NH2+] = x = 0.0172 M

5 (b) Recall that pH + pOH = 14, which can be rearranged to give: pH = 14 – pOH We need the value for pOH: pOH = -log10[−OH] = -log10[0.0172] = 1.76 ∴ pH = 14 – 1.76 = 12.24 = 12.2 7. Hydrofluoric acid (HF) is a weak acid with Ka = 7.2 × 10−4 at 25 °C. For a 0.45 M solution of HF at 25 °C, determine:

(a) [H3O+] (b) pOH (c) percent dissociation of HF (d) Kb for F−

SOLUTION: This is a “weak acid” problem. Let’s start with the chemistry: HF + H2O F− + H3O+ Initial conc. 0.45 M 0 (10−7 M) Change -x +x +x Equilibrium conc. 0.45-x x x

Ka = [F−][ H3O+] = 7.2 × 10−4 [HF]

(x)(x) = 7.2 × 10−4 (0.45-x)

Assume x << 0.45, and thus (0.45-x) ≈ 0.45

x2 = (7.2 × 10−4)(0.45) x = 0.018 M

Check: 0.018 × 100% = 4.0% <5%, assumption is okay 0.045

Now we can answer the parts of the question:

(a) [H3O+] = x = 0.018 M

(b) pOH we can determine from pH:

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pH = -log10[H3O+] = -log10(0.018) = 1.7 since pH + pOH = 14, rearrange to get pOH = 14 – pH ∴ pOH = 14.0 – 1.7 = 12.3

(c) percent dissociation of HF

This is the same calculation we do when we check our assumption. In other words, compare the concentration of what you get at equilibrium (H3O+, in this case,) to the concentration of what you started with (HF).

[H3O+]equilibrium × 100% = 0.018 × 100% = 4.0% dissociated

[HF]initial 0.045

(d) Kb for F−

Recall the equation that relates Ka and Kb for a conjugate acid-base pair: Ka × Kb = Kw which can be rearranged to Kb = Kw Ka Thus, Kb = 1.0 × 10−14 = 1.4 × 10−11 7.2 × 10−4

8. A student dissolves 2.15 g of hydrazoic acid, HN3, to form one litre of solution. The pH of the solution is measured and found to be 3.01. What is the ionization constant of hydrazoic acid? SOLUTION: This is a “weak acid” problem. Start with the chemistry: HN3 + H2O H3O+ + N3

− Ka = [H3O+] [N3

−] [HN3] Ka is unknown, but the pH of the solution is 3.01, and so we can determine the [H3O+]: [H3O+] = antilog(-pH) = 10−pH = 10−3.01 = 9.77 × 10−4M

7 Also, we can determine the initial concentration of the solution: 2.15 g = 0.0500 mol 43.0 g/mol and this is in one litre of solution, so, initially [HN3] = 0.0500 M HN3 H3O+ N3

− Initial conc. 0.0500 M 0 (10−7 M) Change -x +x +x Equilibrium conc. 0.0500-x x x But we calculated [H3O+], and so x = 9.77 × 10−4M = [H3O+] = [N3

−] and [HN3] = 0.0500 - x = 0.0490 M Ka = (9.77 × 10−4)( 9.77 × 10−4) = 1.95 × 10-5 0.0490