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Chemistry Question on Ionic Equilibrium In Solution

If 0.1 M0.1\, M of a weak acid is taken and its percentage of degree of ionization is 1.34%1.34 \%, then its ionization constant will be :

A

0.8×10−5 0.8\times 10^{-5}

B

1.79×10−5 1.79\times 10^{-5}

C

0.182×10−5 0.182\times 10^{-5}

D

None of these

Answer

1.79×10−5 1.79\times 10^{-5}

Explanation

Solution

According to Ostwald's dilution law Ka=Cα21−αK_{a}=\frac{C \alpha^{2}}{1-\alpha} where, Ka=K_{a}= equilibrium constant for weak acid C=C = concentration α=\alpha= degree of ionisation of weak acid (HA)(HA) For very weak acid α<<<<1\alpha<<<<1 or 1−α≃11-\alpha \simeq 1 Ka=Cα2K_{a}=C \alpha^{2} or α=KaC\alpha=\sqrt{\frac{K_{a}}{C}} C=0.1 M,α=1.34%=0.0134C=0.1\, M, \alpha=1.34\% =0.0134 Ka=0.1×(1.34×10−2)2K_{a}=0.1 \times\left(1.34 \times 10^{-2}\right)^{2} =1.79×10−5=1.79 \times 10^{-5}