What is the pH of a 0.1938 mol/L NH3 solution? Kb(NH4OH)=1.790x10^-5.

The pH of the solution of NH4OH with concentration 0.1938 mol/L is 11.27.

Expert Answers

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Ka and Kb refer to the acid and base dissociation constants of a compound. These indicate the extent to which an acid or a base dissociates. A strong acid has a large Ka value, and a strong base has a large Kb value.

The Kb of NH4OH is 1.790 x 10^-5.

For a base `[OH^-]= sqrt(Kb * C)`

As the given NH4OH solution has a concentration of ​0.1938 mol/L and Kb = 1.790 x 10^-5
`[OH^-] = sqrt(1.790 * 10^-5 * 0.1938)`
`[OH^-] = 1.8625*10^-3`
`K_w​=[H^+][OH^-]`
`10^-14=[H^+][1.8625*10^-3]`
`[H^+]= (10^-14)/(1.8625*10^-3)`
`[H^+] = 5.369*10^-12`
The pH of a solution is the negative logarithm of the hydrogen ion concentration `[H^+]` . `pH = -log[H^+]` .
For the given solution, `pH = -log(5.369*10^-12)` = 11.27.
A 0.1938 mol/L NH3 solution, where Kb(NH4OH)=1.790x10^-5 is 11.27.
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