Question

Calculate the pH of 3.0 M HP Ka=6.8 x 10^-4

Answer 1

Let α be the dissociation of the weak acid
                            HP <---> H ⁺ + P^-

initial conc.            c               0         0

change                -cα            +cα      +cα

Equb. conc.         c(1-α)          cα    cα

Dissociation constant , K_a = cα x cα / ( c(1-α)

                                         = c α² / (1-α)

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

So K_a = cα²

==> α = √ ( K_a / c )

Given K_a = 6.8x10^-4

          c = concentration = 3.0 M

Plug the values we get α =0.015

[H⁺] = cα = 3.0 x 0.015 = 0.045 M

pH = - log [H⁺]

    = - log(0.045)

= 1.34