Question

Calculate pH of $1MNaHC{O}_3.$ Given $H_{2}C{O}_3+H_{2}O\rightleftharpoons HC{O}_3^{-}+H_3{O}^{+};p{K}_1=6.38$ $HC{O}_3^{-}+H_{2}O\rightleftharpoons C{O}_3^{2-}+H_3{O}^{+};p{K}_2=10.26$

• A. 8.73
• B. 8.32,
• C. 6.73
• D. 6.32

Answer 1

Answer: (B)

${\text{HCO}}_3^{-}$
is proton acceptor as well as proton donor and therefore it is amphiprotic. For such case
$pH=\frac{p{k}_1+p{k}_2}{2}$
$HC{O}_3^{-}$
undergoes hydrdysis and form alkaline solution.
$HC{O}_3^{-}+H_{2}O{H}_{2}C{O}_3+O{H}^{-}$ $HC{O}_3^{-}$
can also ionise to form acidic solution.
$HC{O}_3^{-}+H_{2}O{H}_3{O}^{+}+C{O}_3^{2-}$ pH of the salt is,
$pH=\frac{p{k}_1+p{k}_2}{2}$
$=\frac{6.38+10.26}{2}=8.32$