Chemistry 65.100 A and V
Fourth Test
March 6, 1998
ANSWERS

Part A.

1. A: Solid
B: Liquid
C: Gas
D: Supercritical fluid
E: 100^oC (or the normal boiling point)

2. Raoult’s law: p = p^o_solv X_solv, where p is the vapor pressure of the solution, p^o_solv is the vapor pressure of the pure solvent, and X_solv is the mole fraction of the solvent in the solution.

3. Since NH₃ is a weak base, the NH₄⁺ ion is acidic. Since HClO₄ is a strong acid, the ClO₄^- ion is neutral. Thus, a solution of NH₄ClO₄ will be acidic.

4. Cl is a more electronegative element than Br. Thus, the Cl atom in HOCl pulls more electron density away from the H-O bond in the molecule, allowing the H ion to dissociate more easily, i.e. HOCl is a stronger acid.

5. HF is the weaker acid of the two. Weak acids have strong conjugate bases. Thus, the F^- ion is a stronger base than the Cl^- ion.

6. When Cu(OH)₂ is put into water, it dissloves, releasing OH^- ions. If we then lower the pH, the OH^- concentration will decrease. The dissolution of Cu(OH)₂ will shift towards more dissolution to replace the OH^-, i.e. higher solubility at lower pH.

Part B.

1. The CO_2(g) will dissolve into the solution to form H₂CO_3(aq):

CO_2(g) + H₂O_(l) ¾ H₂CO_3(aq)

The concentration of H₂CO₃ can be found using Henry’s Law:

[H₂CO_3(aq)] = K_H p_CO2
= 0.032 mol L^-1atm^-1 (3 atm)
= 0.096 mol L^-1

Then the H₂CO₃ dissociates:
H₂CO_3(aq) + H₂O_(l) ¾ H₃O⁺_(aq) + HCO₃^-_(aq)

As usual, set up a table showing the initial and equilibrium concentrations:

here, K_a1 = [H₃O⁺_(aq)][HCO₃^-_(aq)] / [H₂CO_3(aq)]
= x²/(0.096 - x) = 4.4 x 10^-7

Since K_a1 is small, we can assume that 0.096 - x = 0.096. Thus, x²/0.096 = 4.4 x 10^-7. Solving, x = [H₃O⁺] = [HCO₃^-] = 2.06 x 10^-4

Thus, pH = -log₁₀(2.06 x 10^-4) = 3.69

To find [CO₃^-2], use the second dissociation: K_a2 = [H₃O⁺_(aq)][CO₃^-2_(aq)]/[HCO₃^-_(aq)]
Rearranging, [CO₃^-2_(aq)] = K_a2[HCO₃^-_(aq)]/[H₃O⁺_(aq)] = K_a2 (since [H₃O⁺] = [HCO₃^-])
Thus, [CO₃^-2_(aq)] = 5.6 x 10^-11

2. (a) The dissolution reaction is:

SrF_2(s) ¾ Sr⁺²_(aq) + 2 F^-_(aq)

but K_sp = [Sr⁺²_(aq)] [F^-_(aq)]²
If we let x = [Sr⁺²_(aq)], then [F^-_(aq)] = 2x
Thus, K_sp = 2.5 x 10^-9 = x (2x)² = 4x³
Solving, x = [Sr⁺²_(aq)] = 8.55 x 10^-4 mol L^-1

(b) If [NaF] = 0.5 mol L^-1, then [F^-] = 0.5 mol L^-1 also.
But, K_sp = [Sr⁺²_(aq)] [F^-_(aq)]²
Thus, [Sr⁺²] = K_sp / [F^-]² = 2.5 x 10^-9 / (0.5)² = 1.00 x 10^-8 mol L^-1

(c) Since the density is the same as that of pure water, molality = molarity.

DT_b = K_b m = K_bM = 0.51 ^oC kg mol^-1 (1.0 mol L^-1) = 0.51 ^oC
Thus, T_b = 100 + 0.51 = 100.51^oC

(Note that we had to use M = 1.0 M, since NaF dissociates into two ions, and since the concentration of Sr⁺² ions is negligibile compared to the concentration of NaF.)

(d) Osmotic pressure, P = MRT = 1.0 mol L^-1 (0.082 L atm K^-1mol^-1)(298K) = 24.4 atm

3. At 34.1^oC, the vapor pressure of pure water is 40.1 mm Hg. DH_vap for water is 40.7 kJ mol^-1. Calculate the vapor pressure of water at 85.5^oC.

The Clausius-Clapeyron equation is:

ln p₂ = ln p₁ + (DH_vap/R) (1/T₁ - 1/T₂)

here, p₁ = 40.1 Torr, T₁ = 34.1^oC = 307.2 K, T₂ = 85.5^oC = 358.6 K
Thus, ln p₂ = ln(40.1) + (40700 J mol^-1/(8.314 J K^-1 mol^-1) (1/307.2 K - 1/358.6 K)
= 5.98

Thus, p₂ = e^5.98 = 393.7 mm Hg (= 0.52 atm)