• # question_answer For an aqueous solution of $N{{H}_{4}}Cl,$ prove that $[{{H}_{3}}{{O}^{+}}]\,=\sqrt{{{K}_{h}}C}$.

Refer to page 7/136. For salt of strong acid and weak base, $[{{H}_{3}}{{O}^{+}}]\,=\sqrt{\frac{{{K}_{w}}C}{{{K}_{b}}}}$ Substituting $\frac{{{K}_{w}}}{{{K}_{b}}}\,={{K}_{h}},$ we get $[{{H}_{3}}{{O}^{+}}]\,=\sqrt{{{K}_{h}}C}$