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Q.
Calculate the $pH$ of aqueous solution of $1.0\, M \,HCOONH _{4}$ assuming complete dissociation $(p K_{a}$ of $HCOOH =3.8$ and $p K_{b}$ of $NH _{3}=4.8 )$
Equilibrium
Solution:
Ammonium formate undergoes hydrolysis as
$NH _{4}^{+}+ HCOO ^{-}+ H _{2} O \rightleftharpoons NH _{4} OH + HCOOH$
$K_{h}=\frac{K_{w}}{K_{a} \cdot K_{b}}$
Moreover in the solution we have
$\left[ NH _{4} OH \right]=[ HCOOH ]$
Hence $ K_{h}=\frac{[ HCOOH ]^{2}}{\left[ HCOO ^{-}\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 $2 pH =p K_{W}+p K_{a}-p K_{b}$
or $pH =\frac{1}{2}\left[p K_{W}+p K_{a}-p K_{b}\right]$
$=\frac{1}{2}[14+3.8-4.8]=6.5$