Question

Consider 0.50 M HF (K_a = 7.2 times 10^-4). What are the major species in the solution? Calculate the pH of this solution. pH =

Answer 1

Let a be the dissociation of the weak acid,HF
                            HF <---> H ⁺ +     F^-

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 = 7.2x10^-4

          c = concentration = 0.50 M

Plug the values we get a = 0.0379

[H⁺] = ca = 0.50 x 0.0379 M = 0.0189 M

pH = - log[H⁺]

     = - log 0.0189

     = 1.72

Therefore the pH of the solution is 1.72

HF + H₂O <---> H₃O ⁺ + F^-

Since HF is weak acid it will dissociate partially, so the major species in the solution is HF, H⁺, F^-