Question

If the Ka of a monoprotic weak acid is 2.1 × 10-6, what is the pH of a 0.43 M solution of this acid?

Answer 1

Let a be the dissociation of the weak acid
                            HA <---> H ⁺ + A^-

initial conc.            c               0         0

change                -ca            +ca      +ca

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 a is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given K_a = 2.71x10^-6

          c = concentration = 0.43 M

Plug the values we get a = 2.51x10^-3

[H⁺] = ca = 0.43 x 2.51x10^-3 = 1.08x10^-3 M

pH = - log[H⁺]

     = - log (1.08x10^-3 )

     = 2.96

∴ pH of the solution is 2.96