Question

Calculate the pH of an aqueous solution of $1.0M$ ammonium formate assuming complete dissociation. ($p_{Ka}$ of formic acid = 3.8 and $p_{Kb}$ of ammonia = 4.8)

• A. 7.2
• B. 7.5
• C. 6
• D. 6.5

Answer 1

The correct option is D

6.5

This is a bit of a strange situation isn't it?

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

So here we have a mixture of a weak acid and a weak base. So we'll have to use the corresponding hydrolysis equation. Before that, we make a note that $p_{Ka}$ of formic acid = 3.8 and $p_{Kb}$ of ammonia = 4.8 are given.

This makes it easier guess because if we separately write the hydrolysis equations for the weak acid and the weak base, we might figure out a way to get $\left[H^{+}\right]$, which is in a way, what we are looking for

$K_h$ = $\frac{\left[N{H}_{4}OH\right]\left[HCOOH\right]}{\left[N{H}_4^{+}\right]\left[HCO{O}^{-}\right]}$ ............(1)

$K_a$ = $\frac{\left[HCOO\right]\left[H^{\prime }\right]}{\left[HCOOH\right]}$ ............(2)

$K_b$ = $\frac{\left[OH\right]\left[N{H}_4^{\prime }\right]}{\left[N{H}_{4}OH\right]}$ ............(3)

$K_h$ = $\frac{\left[OH\right]\left[H^{\prime }\right]}{K_b.K_a}$ = $\frac{K_w}{K_a{K}_b}$ ............(4)

Moreover, in the very first Ammonium Formate hydrolysis reaction,0

$\left[N{H}_{4}OH\right]$ = $\left[HCOOH\right]$ and $\left[N{H}_4^{+}\right]$ = $\left[HCO{O}^{-}\right]$

Substituting the above in Eq. (1)

$K_h$ = $\frac{[HCOOH{]}^2}{[HCO{O}^{-}{]}^2}$

Using the above and combining (2) and (4), we have

$\frac{K_w}{K_a{K}_b}$ = $\frac{[H^{\prime }{]}^2}{K{a}^2}$

Or $[H^{+}{]}^2$ = $\frac{K_a{K}_w}{Kb}$

Taking $-Lo{g}_{10}$ on both sides of the above equation, we have

$2pH$ = $p_K+p_{Ka}-p_{Kb}$ = $[14+3.8-4.8]$

Or $pH$ = $6.5$