Question

Solving for the pH in a mixture of acids is like dealing with a diprotic acid. You solve the problem by dealing with the acids in successive order. You start with the stronger acid (just like you start with the Ka1 because it is higher than the Ka2 for a diprotic). Afterward, you move on to the weaker acid. If a solution contains 0.029 M HCl and 0.100 M HF (Ka=6.6x10-4), what will be the pH of the solution?

Answer 1

Answer:

1.43 will be the pH of the solution.

Explanation:

HCl

Concentration of HCl = 0.029 M

Concentration of hydrogen ions = $[H^+]$

$[HCl]=[H^+]=0.029 M$

HF

Concentration of HF = c = 0.100 M

Dissociation constant of HF = $K_a=6.6\times 10^{-4}$

$HF\rightleftharpoons H^++F^-$

Initially

c         0          0

At equilibrium

(c-x)     x           x

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][F^-]}{[HF]}=\frac{x\times x}{(c-x)}$

$6.6\times 10^{-4}=\frac{x^2}{(0.100M-x)}$

Solving for x:

x = 0.0078 M

$[H^+]'=x=0.0078 M$

Total hydrogen ion concentration = $[H^+]_t=[H^+]+[H^+]'=0.029 M + 0.0078 M=0.0368 M$

The pH of the mixture with HCl and HF

$pH=-\log[H^+]_t$

$=-\log[0.0368 M]=1.43$

1.43 will be the pH of the solution.