C6H5OH ⇌ C6H5O– + H+
Initial 0.05M 0 0
After dissociation 0.05-x , x, x
\( K_{}\alpha = \dfrac{x^{2}}{0.05-x} \)
Because the value of the ionization constant is very small, the value of x is very small. Accordingly, w e may ignore x in the denominator.
\( x = \sqrt{1\times 10^{-10}\times 0.05} \)
x =
x = 2.2×10-6M
In presence of 0.01 sodium phenolate(C6H5Na), suppose y is the amount of phenol dissociated, then at equilibrium
[C6H5OH] = 0.05 -y
[C6H5OH] ≈ 0.05
[C6H5O–] = 0.01 + y ≈ 0.01M,
[H+]=yM
\( K_{\alpha }=\dfrac{(0.01)(y)}{0.05} \)
\( =1.0\times 10^{-10}\)
y = 5 × 10-10
\( \alpha = \dfrac{y}{c}α= \dfrac{5\times 10^{-10}}{5\times 10^{-2}} \)
α = 10-8
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