We have
[HF] = 0.400 M
Ka = 6.8 10-4
Lets see the dissociation reaction for HF
HF H+(aq) + F-(aq)
Lets see the ICE table for this equilibrium reaction
.............................HF 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 10-4 = (x)(x) / (0.400-x) = x2 / (0.400-x)
As x << 0.400 , 0.400-x = 0.400
then
x2 = 0.400 6.8 10-4 = 2.72 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
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