The \(\mathrm{pH}\) of \(0.1 \mathrm{M} \mathrm{NaHCO}_{3}\) solution can be computed from the expression

The \(\mathrm{pH}\) of \(0.1 \mathrm{M} \mathrm{NaHCO}_{3}\) solution can be computed from the expression
  1. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a} 1}^{\circ}+K_{\mathrm{a} 2}^{\circ}\)
  2. \(\mathrm{pH}=\frac{1}{2}\left(\mathrm{p} K_{\mathrm{a} 1}^{\circ}+\mathrm{p} K_{\mathrm{a} 2}^{\circ}ight)\)
  3. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a} 1}^{\circ}-K_{\mathrm{a} 2}^{\circ}\)
  4. \(\mathrm{pH}=\frac{1}{2}\left(\mathrm{p} K_{\mathrm{a} 1}^{\circ}-K_{\mathrm{a} 2}^{\circ}ight)\)

Solution

\(\mathrm{Na}^{+}\) does not show hydrolysis reaction. \(\mathrm{HCO}_{3}^{-}\) is an amphiprotic anion.
\(\begin{array}{l}
\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} ightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CO}_{3}^{2-} ; \quad K_{\mathrm{a} 2}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}ight]\left[\mathrm{CO}_{3}^{2-}ight]}{\left[\mathrm{HCO}_{3}^{-}ight]} \\
\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} ightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{OH}^{-} ; \quad K_{\mathrm{b}}=\frac{\left[\mathrm{H}_{2} \mathrm{CO}_{3}ight]\left[\mathrm{OH}^{-}ight]}{\left[\mathrm{HCO}_{3}^{-}ight]} \\
\text {Also } K_{\mathrm{b}}=\frac{\left[\mathrm{OH}^{-}ight]\left[\mathrm{H}^{+}ight]}{\left[\mathrm{HCO}_{3}^{-}ight]\left[\mathrm{H}^{+}ight] /\left[\mathrm{H}_{2} \mathrm{CO}_{3}ight]}=\frac{K_{\mathrm{w}}}{K_{\mathrm{a} 1}}
\end{array}\)
From the expression of \(K_{\mathrm{a} 2}\), we get
\({\left[\mathrm{H}_{3} \mathrm{O}^{+}ight]=\frac{K_{\mathrm{a} 2}\left[\mathrm{HCO}_{3}^{-}ight]}{\left[\mathrm{CO}_{3}^{2-}ight]}=\frac{K_{\mathrm{a} 2}\left\{\left[\mathrm{H}_{2} \mathrm{CO}_{3}^{-}ight]\left[\mathrm{OH}^{-}ight] / K_{\mathrm{b}}ight\}}{\left[\mathrm{CO}_{3}^{2-}ight]}=\frac{K_{\mathrm{a} 2}}{\left[\mathrm{CO}_{3}^{2-}ight]} \frac{\left[\mathrm{H}_{2} \mathrm{CO}_{3}ight]}{\left(K_{\mathrm{w}} / K_{\mathrm{a} 1}ight)} \frac{K_{\mathrm{w}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}ight]}}\)
or \({\left[\mathrm{H}_{3} \mathrm{O}^{+}ight]^{2}=K_{\mathrm{a} 1} K_{\mathrm{a} 2} \frac{\left[\mathrm{H}_{2} \mathrm{CO}_{3}ight]}{\left[\mathrm{CO}_{3}^{2-}ight]}}\)
Since both \(K_{\mathrm{a} 2}\) and \(K_{\mathrm{b}}\) are usually very small, one may assume \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}ight] \simeq\left[\mathrm{CO}_{3}^{2-}ight]\). Hence
or \(\left[\mathrm{H}_{3} \mathrm{O}^{+}ight]^{2}=K_{\mathrm{al}} K_{\mathrm{a} 2}\) or \(\mathrm{pH}=\frac{1}{2}\left(\mathrm{p} K_{\mathrm{a} 1}^{\circ}+\mathrm{p} K_{\mathrm{a} 2}^{\circ}ight)\) undefined

Asked in: JEE-TOPICTESTS-CHEMISTRY

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