Question

Find the pH of a 0.230 M HF solution.

Find the percent dissociation of a 0.230 M HF solution.

Answer 1

HF(aq)+H2O(l)⇌H3O+(+F−(
I...0.230 M.................................0....................0
C..(-x).....................................(+x).................(+x)
E..(0.230 -x)...............................x.....................x

According to the definition of the acid dissociation constant,

x⋅x / 0.230 −=x/ 0.230 −=7.2⋅10−4

Solving for x will get you x=0.0128

This is the concentration of both the hydronium, and of the fluride ions in solution.

pH = - log [H3O+]

= - log 0.0128

= 1.89

= 1.90

Percent ionization is calculated by taking the concentration of hydronium ions, dividing it by the initial concentration of the acid and multiplying the ratio by 100.

% ionization=0.0128 M / 0.230 M⋅100

= 5.565 %