Chemistry 65

Chemistry 65.100 A and V
Fourth Test

March 2, 2001
Answers

Part A.

1. A - solid
B - liquid
C - gas
D - solid, gas
E - solid, liquid, gas
F - solid, liquid
G - liquid, gas
E is the triple point
H is the critical point
At F, T = 0^oC
At G, T = 100^oC

2. A solution of NH₄Cl₍_aq) is acidic because the NH₄⁺_(aq) ion is the conjugate acid of a weak base.

When NH₄Cl is dissolved in water, the following equilibrium is set up:

NH₄⁺₍_aq) + H₂O_(l) ¾ NH_3(aq) + H₃O⁺_(aq)

which results in the production of the hydronium ion, i.e. the solution is acidic.

3. Freshwater is naturally acidic because CO₂ from the atmosphere dissolves in the water to produce carbonic acid, a weak acid.

The relevant reactions need not be shown, but are:

CO₂₍_g) + H₂O_(l) ¾ H₂CO_3(aq)
H₂CO_3(aq) + H₂O_(l) ¾ H₃O⁺_(aq) + HCO₃^-_(aq)

4. The intermolecular forces in liquid Xe are greater than those in liquid Ar, since Xe is a larger atom, and is therefore more polarizable.

5. Since the H atom is bound to the O atom in these acids, the strength of the O-H bond is affected by the electronegativity of the other atom. In this case, Cl is more electronegative than Br. Thus, the Cl atom pulls the electrons towards itself strongly, resulting in lower electron density in the O-H bond, hence a weaker bond and a stronger acid.

6. The relevant reactions are:

CaF_2(s) ¾ Ca⁺²_(aq) + 2 F^-_(aq)
HF_(aq) + H₂O_(l) ¾ F^- + H₃O⁺_(aq)

Since HF is a weak acid, the position of the second equilibrium is influenced by the pH. In particular, raising the pH will decrease [H₃O⁺₍_aq)]. The second equilibrium will then shift to the right to increase [H₃O⁺₍_aq)], which will also increase [F^-_(aq)]. The first equilibrium will therefore shift to the left to decrease [F^-_(aq)], i.e. the solubility of CaF_2(s) will decrease.

Part B.

1(a) Molality = moles solute / kg solvent

2.00 L of solution with a density of 1.25 g/mL = 2,000 mL x 1.25 g/mL = 2,500 g solution.
3.00 mol CaCl_2(s) = 3.00 mol x 111.1 g/mol = 333.3 g CaCl_2(s)
Thus, mass solvent = 2,500 g - 333.3 g = 2166.7 g = 2.167 kg
Molality = moles solute / kg solvent = 3.00 mol/2.167 kg = 1.38 mol/kg = 1.38 m CaCl₂ (= 4.14 m in ions)

(b) DT_f = K_f m_solute
= 1.86 ^oC kg mol^-1 (3 x 1.38 mol kg^-1) = 7.70^oC
Thus, the freezing point is 0.0 - 7.70 = -7.70^oC

(c) DT_b = K_b m_solute
= 0.51 ^oC kg mol^-1 (3 x 1.38 mol kg^-1) = 2.11 ^oC
Thus, the freezing point is 100.0 + 2.11 = 102.1^oC

(d) P = MRT
= (9.00 mol/2.000L)(0.082 L atm K^-1 mol^-1)(298 K)
= 110.0 atm

(e) p_solution = p^o_solvent X_solvent
moles ions = 3.00 x 3 = 9.00
moles water = 2,167 g / 18 g mol^-1 = 120.4
Thus, X_solvent = 120.4/(9.00+120.4) = 0.930
Thus, p_solution = 23.8 mm Hg x 0.930 = 22.1 mm Hg

2. This is an application of the Clausius-Clapeyron equation:

ln(p₂/p₁) = DH_vap/R [1/T₁ - 1/T₂]

Solving for p₂, we have:

ln p₂ = DH_vap/R [1/T₁ - 1/T₂] + ln p₁

Note that p₁ is the vapor pressure of the liquid at its normal boiling point, i.e. p₁ = 1 atm

thus, ln p₂ = 35760 J mol^-1 / 8.314 J K^-1 mol^-1 [1/(273.1+99.2 K) - 1/(273.1 + 25 K)] + ln(1)
= -2.88

Thus, p₂ = e^-2.88 = 0.0564 atm

(= 42.8 mm Hg if p₁ = 760 mm Hg was used)

3(a) The base hydrolyzes according to:

NH₃₍_aq) + H₂O_(l) ¾ NH₄⁺_(aq) + OH^-_(aq)

and the salt dissociates fully:

NH₄Cl_(s) ® NH₄⁺₍_aq) + Cl^-_(aq)

To find the pH, we use the Henderson-Hasselbalch equation:

pH = pK_a - log₁₀[acid]/[base]

Where [acid] = [NH₄⁺₍_aq)] = 0.40 M (most of the ammonium ion comes from dissociation of the salt)

and [base] = [NH_3(aq)] = 0.25 M (very little of this weak base has dissociated)

pK_a refers to the acid in the system, i.e. to the NH₄⁺_(aq). To find its pK_a, we use the relationship between K_a and the K_b of its conjugate base, NH₃:

K_a = K_w / K_b
= 1.0 x 10^-14 / 1.8 x 10^-5
= 5.56 x 10^-10

Thus, pK_a = -log₁₀ K_a
= -log₁₀ (5.55 x 10^-10)
= 9.25

Thus, pH = 9.25 - log₁₀(0.40/0.25)
= 9.05

(b) When the acid is added to the solution, it consumes some of the base according to:

HCl₍_aq) + NH_3(aq) ® NH₄⁺_(aq) + Cl^-_(aq)

Thus, the base concentration decreases to 0.25 M - 0.05 mol/L = 0.20 M
And the conjugate acid concentration increases to 0.40 M + 0.05 mol/L = 0.45 M

And so the pH is now:
pH = 9.25 - log₁₀(0.45/0.20)
= 8.90

4. When the sulfurous acid is dissolved in water, the following equilibria are established:

H₂SO₃₍_aq) + H₂O_(l) ¾ H₃O⁺_(aq) + HSO₃^-_(aq) K_a1 = 0.015
HSO₃^-_(aq) + H₂O_(l) ¾ H₃O⁺_(aq) + SO₃^-2_(aq) K_a2 = 6.3 x 10^-8

[H₃O⁺][HSO₃^-]/[H₂SO₃] = 0.015

At equilibrium, x(x)/(1.00 - x) = 0.015
Since K_a1 > 10^-3, we can not make the simplifying assumption that x<<1.00, and so this must be solved using the quadratic equation.
x² = 0.015 (1.00 - x)
x² + 0.015 x - 0.015 = 0

Solving, x = 0.115 or -0.13
Only the positive root makes any sense since x is a concentration.

Thus, x = [H₃O⁺] = [HSO₃^-] = 0.115 M
[H₂SO₃] = 1.00 - x = 0.885 M

Since K_a1 >> K_a2, we know that essentially all of the H₃O⁺ will come from the first disociation. Thus,

pH = -log₁₀[H₃O⁺] = -log₁₀(0.115) = 0.94

From the second dissociation,
[H₃O⁺][SO₃^-2]/[HSO₃^-] = 6.3 x 10^-8
Note that the values of [H₃O⁺] and [HSO₃^-] come from the first dissociation, (and are equal to one another). Thus,

0.115[SO₃^-2]/ 0.115 = 6.3 x 10^-8
or, [SO₃^-2] = 6.3 x 10^-8 M