Question

The weak acid HZ has a Ka of 2.55 x 10-4. Calculate the pH of a 0.235 M solution.

Answer 1

Let a be the dissociation of the weak acid
                            HZ      H ⁺ + A^-

initial conc.            c               0         0

Equb. conc.         c(1-a)          ca       ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

                                         = c a² / (1-a)

In the case of weak acids α is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given K_a = 2.55x10^-4

          c = concentration = 0.235 M

Plug the values we get a = 0.0329

[H⁺] = ca

       = 0.235 x 0.0329 M

      = 7.74x10^-3 M

pH = - log [H⁺]

     = - log (7.74x10^-3)

    = 2.1

Therefore the pH of the solution is 2.1