Question

Calculate the pH of a 0.100M NaCH₃COO solution. Ka for acetic acid, HC₂H₃O₂ is 1.8x10^-5

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 = 1.8x10^-5

          c = concentration = 0.100 M

Plug the values we get a = 0.0134

[H⁺] = ca = 0.100 x 0.0134 M

       = 1.34x10^-3 M

pH = -log[H⁺]

     = - log ( 1.34x10^-3)

     = 2.87

Therefore the pH of the solution is 2.87