The equlibrium constant of the reaction of a weak acid (HA) with a strong base $(\mathrm{NaOH}$) is $10^{9}$…

The equlibrium constant of the reaction of a weak acid (HA) with a strong base $(\mathrm{NaOH}$) is $10^{9}$ at $25^{\circ} \mathrm{C}$. Hence $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{A}^{-}$ will be equal to
  1. $10^{-5}$
  2. $10^{-9}$
  3. $10^{5}$
  4. $10^{-6}$

Solution

$\mathbf{H A} ightleftharpoons \mathbf{H}^{+}+\mathbf{A}^{-}, \mathbf{K}$
$\mathbf{H}^{+}+\mathbf{O H}^{-} ightleftharpoons \mathrm{H}_{2} \mathrm{O}, \frac{1}{\mathbf{K}_{\mathrm{w}}}$
$\mathrm{HA}+\mathrm{OH}^{-} ightleftharpoons \mathrm{H}_{2} \mathrm{O}+\mathbf{A}^{-}, \mathbf{K}_{\mathbf{b}}=\frac{\mathbf{K}_{\mathrm{a}}}{\mathbf{K}_{\mathrm{w}}}=10^{9}$
$\frac{\mathrm{K}_{\mathrm{a}}}{10^{-14}}=10^{9}$ so $\mathrm{K}_{\mathrm{a}}=10^{-5}$
$\therefore \mathbf{K}_{\mathbf{g}} \mathbf{K}_{\mathbf{b}}=\mathbf{K}_{\mathbf{w},}, \mathbf{K}_{\mathbf{b}}=\frac{\mathbf{K}_{\mathrm{N}}}{\mathbf{K}_{\mathbf{n}}}=\mathbf{1 0}^{-\mathbf{9}}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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