Question

The Ka of a monoprotic weak acid is 8.6x 10-6, what is the pH of a 0.38 M solution of this acid?

Answer 1

Lets assume weak acid is HA.

[HA] = 0.38

Weak acid does not dissociate completely so we set up ICE chart and its reaction with water.

            HA(aq)     + H2O (l)       ---- > A- (aq )         +           H3O+ (aq)

I           0.38                                         0                                              0

C         -x                                             +x                                            +x

E          (0.38-x)                                   x                                              x

Ka expression for dissociation of HA

Ka = [H3O+] [ A-] / [HA]

Here all the concentrations are used at equilibrium. Lets plug the value of ka and equilibrium concentrations.

8.6 E-6 = x²/ (0.38-x)

Since value of ka is very small so we can make 5% approximation. In it we can neglect x value in the bracket of right side.

8.6 E-6 = x²/ 0.38

x = 1.81 E-3

[H3O+] = x = 1.81 E-3 M

Now pH = - log [H3O+]

pH = - log (1.81E-3 )

= 2.74

So the pH of this solution would be 2.74