Question

If the freezing point of 0.1 M aq. BOH solution is –0.2046°C, then the pH of solution is (Kf for water = 1.86 K.kg mol–1 and assume molarity = molality)​

Answer 1

pH of the solution = 2

Explanation:

Given: Freezing point of 0.1 M aq. BOH solution = –0.2046°C

Find: pH of solution.

Solution:

Given Kf for water = 1.86 K.kg mol–1 and molarity = molality

ΔTf = molality * Kf * i

i = 1 + (n-1)α

α = 0.1

[H+] = Cα

 [H+] = 0.1 * 0.1  = 10^-2

pH = -log[H+]

pH = 2