If $K_{\mathrm{a} 1}\left(\mathrm{H}_{2} \mathrm{CO}_{3}ight)=4.5 \times 10^{-7} \mathrm{M}$ and $K_{\mathrm{a}}\left(\mathrm{HCO}_{3}^{-}ight)=5.0 \times 10^{-11} \mathrm{M}$, the $\mathrm{pH}$ of $0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ solution at $25^{\circ} \mathrm{C}$ will be

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Aicharya · Accepted Answer

$\mathrm{CO}_{3}^{2-}+\mathrm{H}_{2} \mathrm{O} ightleftharpoons \mathrm{HCO}_{3}^{-}+\mathrm{OH}^{-} ;$ $\mathrm{pH}=\frac{1}{2}\left[\mathrm{p} K_{\mathrm{w}}^{\circ}+\mathrm{p} K_{\mathrm{a}}^{\circ}\left(\mathrm{HCO}_{3}^{-}ight)+\log (c / \mathrm{M})ight]$ $=\frac{1}{2}\left[14-\log \left(5 \times 10^{-11}ight)+\log (0.1)ight]=\frac{1}{2}[14+10.30-1]=11.65$ ,

If \(K_{\mathrm{a} 1}\left(\mathrm{H}_{2} \mathrm{CO}_{3}ight)=4.5 \times 10^{-7} \mathrm{M}\) and…

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