Question

What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that is buffered at a pH of 4.78 and has a freezing point of -2.0 ∘C? (Assume complete dissociation of sodium benzoate and a density of 1.01 g/mL for the solution.)

Express your answers using two significant figures separated by a comma.

answer is not 0.11 and 0.44

answer is not 0.11 and 0.45

answer is not 0.11 and 0.46

Textbook says Tf is 1.86 degrees C

Answer 1

pH of buffer solution is given by Henderson's equation,

pH = pK_a + log [salt]/[acid]

where pK_a is negative logarithm of acid dissociation constant K_a for the weak acid; [acid] = concentration of the weak acid; [salt] = concentration of salt of the weak acid with strong base (here sodium benzoate).

pK_a for ​benzoic acid is 4.20; given pH of buffer solution is 4.78; using Henderson's equation,

pH = pK_a + log [salt]/[acid]

4.78 = 4.20 + log [PhCOONa]/[PhCOOH]       (PhCOONa = sodium benzoate; PhCOOH = benzoic acid)

log [PhCOONa]/[PhCOOH] ​​ = 4.78 - 4.20 = 0.58

[PhCOONa]/[PhCOOH] ​ = antilog 0.58 = 3.802

Hence, [PhCOONa]/[PhCOOH] ​ = 3.802​       .........................(1)

​Freezing point of water is 0^oC; given, freezing point of buffer solution is -2.0^oC;

Hence, depression in freezing point ​T_f = 2.

Depression in freezing point ​T_f​ is given by, ​T_f​ = K_f * m

K_f is molal depression constant = 1.86^o/kg (for water); m is molality ( number of moles of solute in 1 kg of solvent)

So, m = ​T_f​ / K_f   = 2 / 1.86 = 1.075 mol / kg

Hence, molality of buffer solution is 1.075 mol / kg or the buffer solution consists of a total of 1.075 mol of solute particles dissolved in 1kg of water. Since density of water is 1 g/mL or 1 kg/L it follows that 1kg of water corresponds to 1L of water. Therefore, the buffer solution consists of a total of 1.075 mol of solute particles dissolved in 1L of water.

Hence, number of moles of solute particles per litre of solution is 1.075

or total molar concentration of all solute particles is 1.075 mol / L.      ............(2)

In the solution,

PhCOOH ​ PhCOO^- + H⁺   and PhCOONa PhCOO^- + Na⁺

PhCOONa is strong electrolyte and dissociates completely as shown above. Due to presence of common ion,

PhCOO^- , the dissociation of PhCOOH is suppressed and remains largely in undissociated form as PhCOOH. So extent of ionization of PhCOOH in the buffer solution is negligible.

So the solute particles in the buffer solution are PhCOOH, PhCOO^- ​and Na⁺ .

Total molar concentration of solute particles = [PhCOOH] + [PhCOO^- ] + [Na⁺]

Since, 1mol PhCOONa dissociates into 1mol PhCOO^- and 1mol Na⁺,

[PhCOO^- ] + [Na⁺]​ = 2 [PhCOONa]

Hence, Total molar concentration of solute particles = [PhCOOH] + 2 [PhCOONa]

Using result (2),

Total molar concentration of solute particles = [PhCOOH] + 2 [PhCOONa] ​= 1.075 mol / L

Hence, [PhCOOH] + 2 [PhCOONa] ​= 1.075 mol / L         ...................(3)

From result (1), [PhCOONa]/[PhCOOH] ​ = 3.802    or   [PhCOONa] ​= 3.802 * [PhCOOH]

Substituting for [PhCOONa] ​​in result (3),

[PhCOOH] + 2 * 3.802 * [PhCOOH]​ = 1.075     or   8.604 * [PhCOOH] = 1.075

Hence, [PhCOOH] ​= 0.125 mol /L

Substituting for [PhCOOH] ​​in result (1),

[PhCOONa]/[PhCOOH] ​ = 3.802    or [PhCOONa] / 0.125 = 3.802

[PhCOONa] = 3.802 * 0.125 = 0.475 mol / L

Hence, [PhCOONa] ​= 0.475 mol / L

Since, the answers are to be expressed using two significant figures,

molar concentration of benzoic acid [PhCOOH] ​= 0.125 mol /L 0.12 mol / L

molar concentration of ​sodium benzoate [PhCOONa] ​= 0.475 mol / L 0.48 mol / L

ANSWER: 0.12, 0.48