Question

The expression to compute pH of $N{H}_{4}Cl$ solution is (at 298 K)

• A. pH = 1/2 pK_w + 1/2 pK_b(NH₄OH) + 1/2 log[salt]
• B. pH = 1/2 pK_w + 1/2 pK_b(NH₄OH) - 1/2 log[salt]
• C. pH = 1/2 pK_w - 1/2 pK_b(NH₄OH) - 1/2 log[salt]
• D. pH = 1/2 pK_w - 1/2 pK_b(NH₄OH) + 1/2 log[salt]

Answer 1

The correct option is C

pH = 1/2 pK_w - 1/2 pK_b(NH₄OH) - 1/2 log[salt]

$N{h}_4^{+}+H_{2}O\rightleftharpoons N{H}_{4}OH+H^{+}$

$Kh$ = $\frac{K_w}{K_b}$ = $\frac{\left[N{H}_{4}OH\right]\left[H^{\prime }\right]}{\left[N{H}_4^{+}\right]}$ = $\frac{[H^{\prime }{]}^2}{\left[N{H}_4^{+}\right]}$

We take $-lo{g}_{10}$ on both sides and what we have is pH = $\frac{1}{2}p{K}_w-\frac{1}{2}p{K}_b\left(N{H}_{4}OH\right)-\frac{1}{2}log\left[salt\right]$