Ncert Exercises & Solutions

7.44 The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

Ionization of phenol:

C_{6} H_{5} OH + H_{2} O \leftrightarrow C_{6} H_{5} O^{-} + H_{3} O^{+}

Initial conc . 0.05 0 0

At equilibrium 0.05 - x x x

K_{a} = \frac{[C_{6} H_{5} O^{-}] [H_{3} O^{+}]}{[C_{6} H_{5} OH]}

K_{a} = \frac{x \times x}{0.05 - x}

As the value of the ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator.

\begin{aligned} ∴ x & = \sqrt{1 \times 10^{- 10} \times 0.05} \\ = \sqrt{5 \times 10^{- 12}} \\ = 2.2 \times 10^{- 6} M = [H_{3} O^{+}] \\ Since [H_{3} O^{+}] = [C_{6} H_{5} O^{-}], \\ [C_{6} H_{5} O^{-}] = 2.2 \times 10^{- 6} M . \end{aligned}

Now, let

\propto

be the degree of ionization of phenol in the presence of

0.01 M C_{6} H_{5} ONa .

C_{6} H_{5} ONa \to C_{6} H_{5} O^{-} + {Na}^{+}

Conc, 0.01

Also,

C_{6} H_{5} OH + H_{2} O \leftrightarrow C_{6} H_{5} O^{-} + H_{3} O^{+}

Conc. 0.05 - 0.05

α

0.05

α

0.05

α

\begin{array}{l} [C_{6} H_{5} OH] = 0.05 - 0.05 α; 0.05 M \\ [C_{6} H_{5} O^{-}] = 0.01 + 0.05 α; 0.01 M \\ [H_{3} O^{+}] = 0.05 α \\ K_{a} = \frac{[C_{6} H_{5} O^{-}] [H_{3} O^{+}]}{[C_{6} H_{5} OH]} \\ K_{a} = \frac{(0.01) (0.05 α)}{0.05} \\ \begin{aligned} 1.0 \times 10^{- 10} & = .01 α \\ α & = 1 \times 10^{- 8} \end{aligned} \end{array}

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