Question

Calculate the pH of the solution of `10ml` of `0.5M` `RNH_3Cl` (`K_h=10^(-9)`) + `40ml` of `0.125M` `KOH`?

Answer 1

`RNH_3^+` is merely an ammonium analog, that acts as an acid, with a chloride counterion.

Consider the equilibrium,

`HRNH_2^(+)+OH^(-) rightleftharpoons H_2O + RNH_2`

Hydroxide ions will neutralize one molar equivalent of protons from this ammonium analog, like so,

From this data we can conclude the reaction is fully neutralized.

However, we still have ammonia analogs in the solution, giving this equilibrium,

`RNH_2 rightleftharpoons RNH_3^(+) + OH^(-)`

where,

`K_"b" = ([RNH_3^+][OH^-])/([RNH_2]) approx 1.8*10^-5`

Analyzing this equilibrium, tabulated below,

Recall, `[OH^-] approx 1.0*10^-7` from `K_"w"`.

Now,

`K_"b" = (x * (1.0*10^-7 + x))/(0.005 - x) = 1.8*10^-5`

`=> x = [OH^-] approx 2.9*10^-4"M"`

Note, since `K_"a" < < K_"b"`, it's safe to assume the acidic dissociation will not significantly affect the `"pH"` I'm about to calculate.

From the hydroxide concentration we can calculate `"pH"` easily. Hence,

`"pH" = 14 + log[OH^-] approx 10.46`