Calculate the value of Ka for this acid.

Question

A 0.250 g sample of the weak acid HA (molar mass = 150.0 g/mol) is dissolved in 50.0 g of water. The freezing point of the resulting solution is -0.071 °C. Calculate the value of Ka for this acid. Assume molality equals molarity in this solution. (Kf for water is 1.86 °C/m.)

Answer 1

To get started, let's first write out the acid dissociation reaction:

HA_(aq) + H₂O(l) ↔ H₃O⁺(aq) + A^-(aq)

This reaction has K_a = ([H₃O⁺]_eq [A^-]_eq)/[HA]_eq

To find these equilibrium concentrations and solve for K_a, we'll need to use the information provided in the question. The question states that after the weak acid is added, the freezing point of the solution drops from 0°C, the freezing point of pure water, to -0.071°C. We can use the freezing point depression equation to solve for i, or the van't Hoff factor, which will tell us how many moles of particles we have in solution per mole of solute:

ΔT_f = -iK_fm_solute where m_solute is the molality of the solute

-0.071°C = -i(1.86°C/m)(0.00167mol HA/.050kg H₂O)

-0.071°C = -i(1.86°C/m)(.0334m)

i = 1.145

This tells us that we have 1.145 moles of particles (HA, H₃O⁺, A^-) per 1 mole of solute (HA) in our solution. For this to be true, 14.5% of HA molecules would have to undergo the dissociation reaction and 85.9% would have to remain undissociated. Let's now use this information to find the equilibrium concentrations to solve for K_a.

Since we can assume that molality equals molarity in this solution, the initial concentration of HA immediately after it is added is 0.0334M. As equilibrium is established, 14.5% of HA will be used to make products, producing [H₃O⁺]_eq = [A^-]_eq = .145(0.0334M) = 0.004843M, while 85.5% of HA will remain as it is, leading to [HA]_eq = .855(0.0334M) = 0.028557M. Plugging these values into the K_a expression, we get:

K_a = ([H₃O⁺]_eq [A^-]_eq)/[HA]_eq = ((0.004843)(0.004843))/(0.028557) = 8.2 x 10^-4