When 2.55 g of an unknown weak acid (MW: 85 g/mol) was added to 250. g of water, the freezing point of the resulting solution is -0.257 °C. What is the Kb of this acid?

Solution:

1) When the acid dissolves it ionizes. What we need to determine is the total amount of solute particles in the solution after it ionizes. We start with the freezing point depression equation:

ΔT = i Kf m

2) Calculate the molality:

2.55 g / 85.0 g/mol = 0.0300 mol

0.0300 mol / 0.250 kg = 0.120 m

3) Calculate the van 't Hoff factor:

0.257 °C = (i) (1.86 °C / m) (0.120 m)

i = 1.151434 <--- I'll keep some guard digits

4) A key assumption is that the density of the solution is 1.00 g/mL. This allows me to treat the 0.120 molal solution as being 0.120 molar.

5) The van 't Hoff factor is reflective of the total number of solute particles in solution, i.e. the total concentration:

(0.120 M) (1.151434) = 0.138172 M<--- rounded off a bit

6) Now, switch to the behavior of the acid. I will call it HA:

HA ---> H+ + A¯

Ka = ([H+] [A¯]) / [HA]

7) We MUST know all the concentrations in order to get the Ka. We can do that using the total concentration of the solute particles:

[H+] + [A¯] + [HA] = 0.138172 M

(x) + (x) + (0.120 - x) = 0.138172

x = 0.018172 M

8) We can now solve for the Ka:

Ka = [(0.018172) (0.018172)] / 0.101828

Ka = 0.003242935

9) We determine the Kb from this relationship:

Ka * Kb = Kw

(0.003242935) (Kb) = 1.00 x 10¯14

Kb = 3.08 x 10¯12