For an acidic buffer containing aceetic acid, `CH_(3)COOH`, and sodium acetate, `CH_(3)COONa`, we have
`[H^(o+)] = (K_(a) ["acid"])/[("salt")]`
(Use this formula, rather than) `pH = pK_(a) + "log" (["salt"])/(["acid"])`
`[CH_(3)COOH] = 0.1M,`
`[H^(o+)] = 10^(-4)` and let `[CH_(3)COONa] = xM`
`rArr [H^(o+)] = 10^(-4) = (2xx10^(-5)xx0.1)/(x)`
`rArr x = 0.02` mol, i.e. `0.02` mol of `CH_(3)COONa` is required.