Question

Calculate pH of $ 1\,M\,NaHC{{O}_{3}}. $ Given $ {{H}_{2}}C{{O}_{3}}+{{H}_{2}}O \rightleftharpoons HCO_{3}^{-}+{{H}_{3}}{{O}^{+}};p{{K}_{1}}=6.38 $ $ HCO_{3}^{-}+{{H}_{2}}O \rightleftharpoons CO_{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} $
$ HCO_{3}^{-} $
undergoes hydrdysis and form alkaline solution.
$ HCO_{3}^{-}+{{H}_{2}}O{{H}_{2}}C{{O}_{3}}+O{{H}^{-}} $ $ HCO_{3}^{-} $
can also ionise to form acidic solution.
$ HCO_{3}^{-}+{{H}_{2}}O{{H}_{3}}{{O}^{+}}+CO_{3}^{2-} $ pH of the salt is,
$ pH=\frac{p{{k}_{1}}+p{{k}_{2}}}{2} $
$ =\frac{6.38+10.26}{2}=8.32 $