Question

a weak acid HX has a dissociation constant 10^-5. the ph of 0.1M solution of this acid will be

Answer 1

HX + H2O <==> H3O+ + X¯
Ka = ( [H3O+] [X¯] ) / [HX] = 10^-5

[H3O+] = x

[X¯] = x

plugging into the equation we have

10^-5 = {(x) (x)} / [HX]

In this problem, the [HX] was originally .1 M and decreased as HX molecules dissociated.
by 'x' amount to become 0.1 - x.

So, 10^-5 = {(x) (x)} / [0.1-x]
x^2 = 10^-5 * (0.1-x)
we can solve the quadratic equation or make the reasonable approximation that x <<<.1
so x^2 = 10^-6
x = 10 ^-03
[H3O=] = 10 ^-03
pH = 3