Question

What is the pH of a 0.400 M. solution of HF (Ka = 6.8 x 10-4)?

Answer 1

We have

[HF] = 0.400 M

Ka = 6.8 $\times$10^-4

Lets see the dissociation reaction for HF

HF $\rightleftharpoons$ H⁺_(aq) + F^-_(aq)   

Lets see the ICE table for this equilibrium reaction

.............................HF   $\rightleftharpoons$ H⁺_(aq) + F^-_(aq)

Initial ...............0.400 M ..........0................0

Change.................-x................+x................+x

At equilibrium.. 0.400-x ...........x.................x

the equilibrium constant for above reaction is

Ka = [H⁺][ F^-] / [HF]

then

6.8 $\times$ 10^-4 = (x)(x) / (0.400-x) = x² / (0.400-x)

As x << 0.400 , 0.400-x = 0.400

then

x² = 0.400 $\times$ 6.8 $\times$ 10^-4 = 2.72 $\times$ 10^-4

by taking square root on both the sides we get

x = 0.016 M

So

[H⁺] = x = 0.016 M

Now we know that

pH = -log[H⁺]

= -log(0.016)

= 1.80

So our answer is 1.80