Question

Calculate the $pH$ of aqueous solution of $1.0MHCOON{H}_4$ assuming complete dissociation $(p{K}_a$ of $HCOOH=3.8$ and $p{K}_b$ of $N{H}_3=4.8)$

• A. 3.5
• B. 6.5
• C. 5.5
• D. 4.5

Answer 1

Answer: (B)

Ammonium formate undergoes hydrolysis as
$N{H}_4^{+}+HCO{O}^{-}+H_{2}O\rightleftharpoons N{H}_{4}OH+HCOOH$
$K_h=\frac{K_w}{K_a\cdot K_b}$
Moreover in the solution we have
$\left[N{H}_{4}OH\right]=[HCOOH]$
Hence $K_h=\frac{[HCOOH{]}^2}{{\left[HCO{O}^{-}\right]}^2}$
or $\frac{K_w}{K_a\cdot K_b}=\frac{{\left[H^{+}\right]}^2}{{\left[K_a\right]}^2}$
or ${\left[H^{+}\right]}^2=\frac{K_w\cdot K_a}{K_b}$
or $2pH=p{K}_W+p{K}_a-p{K}_b$
or $pH=\frac{1}{2}\left[p{K}_W+p{K}_a-p{K}_b\right]$
$=\frac{1}{2}[14+3.8-4.8]=6.5$