Chemistry 65.100 A and V
Fourth Midterm Test Answers
March 5, 1999

Part A

1.  A: Solid; B: Liquid; C: Gas; D: Supercritical Fluid; E: Triple Point; F: Critical Point; G: 0 ^oC; H: 100 ^oC

2. Adding a non-volatile solute decreases the vapor pressure of the solvent. Thus, to make the solution boil requires adding additional heat to increase the vapor pressure back to 1 atm, i.e. we must raise the temperature.

3. KF_(s) is a soluble compound (as are all K salts). However, the F^-_(aq) ions produced will be in equilibrium with HF, a weak acid according to:

F^-_(aq) + H₂O_(l) ¾ OH^-_(aq) + HF_(aq)

Dissolving KF in water will thus increase [F^-_(aq)], which, according to le Chatelier's principle, will shift the above equilibrium to the right, producing  OH^-_(aq), and causing the pH to rise.

4. The extra O atom in HClO₂, being electronegative, pulls more electron density out of the O-H bond, weakening it, allowing the H⁺ ion to be removed more easily, i.e causing HClO₂ to be a stronger acid.

5. The two bases are H₂O and NO₃^-. Since HNO₃ is a strong acid, the dissociation reaction goes essentially to completion, destroying nearly all of the HNO₃ molecules, and producing the equivalent amount of the nitrate ion, NO₃^-. This indicates that H₂O is a stronger base than NO₃^-, since the proton was transferred to it, forming H₃O⁺.

6. The dissolution of Cu₃(PO₄)_2(s) occurs according to:

Cu₃(PO₄)_2(s) ¾ 3 Cu⁺²_(aq) + 2 PO₄^-3_(aq)

But since phosphoric acid, H₃PO₄, is a weak acid, we know that it does not dissociate much, and the following equilibrium is set up:

HPO₄^-2 + H₂O ¾ H₃O⁺ + PO₄^-3

Thus, as the pH is lowered ([H₃O⁺] is increased), the above euilibrium will shift left, consuming some of the PO₄^-3 ions. To make up for this, the first equilibrium will shift right, causing more Cu₃(PO₄)_2(s) to dissolve, i.e. higher solubility at lower pH.

Part B

1a.[15] The Henderson-Hasselbach equation is pH = pK_a + log₁₀{[base]/[acid]}. In this solution, we have acetic acid:

CH₃COOH_(aq) + H₂O_(l) ¾ H₃O⁺_(aq) + CH₃COO^-_(aq)

and so [base] = [CH₃COO^-] = x
and [acid] = [CH₃COOH] = 3.0 M

pK_a = -log₁₀(K_a) = -log₁₀(1.8 x 10^-5) = 4.74
rearranging the above equation, log₁₀[base] = pH - pK_a + log₁₀[acid]
thus, log₁₀[base] = 3.74 - 4.74 + log₁₀[3.0] = -0.52
thus, [base] = [CH₃COO^-] = [sodium acetate] = 10^-0.52 = 0.30 M

1b.[15] The 0.05 mol of base will consume the same amount of acetic acid. Thus, [acid] decreases to 3.0 - 0.05 = 2.95 M. Similarly, the concentration of base (CH₃COO^-) increases by the same amount, to 0.35 M. Then the pH can be calculated by the HH equation:

pH = pK_a + log₁₀{[base]/[acid]}
= 4.74 + log₁₀{0.35 / 2.95}
= 3.81

2. [30] The acid dissociates in water according to:

H₂SeO_3(aq) + H₂O_(l) ¾ H₃O⁺_(aq) + HSeO₃^-_(aq)
and
HSeO₃^-_aq) + H₂O_(l) ¾ H₃O⁺_(aq) + SeO₃^-2_(aq)

Using the first dissociation (from which most of the acidity results), we have:

Thus, at equilibrium,
[H₃O⁺_(aq)] [HSeO₃^-_(aq)]/ [H₂SeO_3(aq)] = K_a1 = 0.035
or,
x²/(2.00 - x) = 0.035
0.035(2-x) = x²
x² + .035x - .070 = 0

Using the quadratic formula, x = 0.25 (or -0.28, which is rejected as a physically meaningless root)

Thus at equilibrium, [H₂SeO₃] = 2.00 - x = 1.75 M
[H₃O⁺_(aq)] = x = 0.25 M
[HseO₃^-_(aq)] = x = 0.25 M

To find [SeO₃^-2], we use the second dissociation:

Thus, at equilibrium,
[H₃O⁺_(aq)] [SeO₃^-2_(aq)]/ [HSeO₃^-_(aq)] = K_a2 = 5.0 x 10^-8
or,
(0.25+x)(x)/(0.25-x) = 5.0 x 10^-8

Since this is such a weak dissociation, we can make the assumption that x<<0.25. Thus the above simplifies to:

x = 5.0 x 10^-8

Thus, [SeO₃^-2_(aq)] = x = 5.0 x 10^-8

Also, pH = -log₁₀[H₃O⁺_(aq)] = -log₁₀(0.25) = 0.60

3a.[15] Since the density of the solution is 1.12 g/mL, the mass of the solution is 1.12 g/mL x 1.00 L = 1120 g. This is made of 1.00 moles Na₂SO₄, which is 1.00 mol x 142.1 g/mol = 142.1 g. The mass of water is therefore 1120 - 142.1 = 977.9 g. The molality of the salt can now be calculated:

m_Na2SO4 = 1.00 mol / 0.9779 kg = 1.02 m

BUT, dissolving Na₂SO_4(s) results in the production of THREE moles of ions per mole Na₂SO_4(s):

Na₂SO_4(s) ® 2 Na⁺_(aq) + SO₄^-2_(aq)

Thus, 1.00 m Na₂SO₄ is really 3(1.02) = 3.06 m in total ions.

DT_f = K_f m
= 1.86 ^oC kg mol^-1 (3.06 mol kg^-1) = 5.7 ^oC
Thus, T_f = 0.00 - 5.6 = -5.7^oC

Similarly,

DT_b = K_b m
= 0.51 ^oC kg mol^-1 (3.06 mol kg^-1) = 1.6 ^oC
Thus, T_f = 100.00 + 1.6 = 101.6^oC

3c.[15] Using Raoult's law, P = P^o_solvent X_solvent.
Here, P^o_solvent = 23.8 mm Hg.
The mole fraction of the solvent can be calculated from the numbers in part a.
moles ions = 3(1.00) = 3.00 moles
moles water = 977.9 g / 18.0 g mol^-1 = 54.33 moles
Thus, X_solvent (= X_water) = 54.33/(3.00+54.33) = 0.95

Thus, P = 23.8 mm Hg (0.95) = 22.6 mm Hg

4a.[15] The relevant reaction is PbCl_2(s) ¾ Pb⁺²_(aq) + 2 Cl^-_(aq). For this reaction,

K_sp = [Pb⁺²_(aq)] [Cl^-_(aq)]²

If x moles of PbCl_2(s) dissolves, we have [Pb⁺²] = x and [Cl^-] = 2x. Thus at equilibrium, we have:

K_sp = 1.2 x 10^-5 = x(2x)² = 4x³

Solving, x = [Pb⁺²] = 0.0144 M

4b.[15] If [Cl^-] = 1.0 mol L^-1, we expect the lead ion concentration to decrease according to the common ion effect:

K_sp = [Pb⁺²_(aq)] [Cl^-_(aq)]²
thus, [Pb⁺²_(aq)] = K_sp / [Cl^-_(aq)]²
= 1.2 x 10^-5 / (1.0)²
= 1.2 x 10^-5 M