Question

0.2(M) solution of monobasic acid is dissociated to $0.95\%$ calculate its dissociation constant.
C = 0.2 M

• A. $1.6\times {10}^{-4}$
• B. $1.8\times {10}^{-5}$
• C. $2\times {10}^{-4}$
• D. $1.7\times {10}^{-5}$

Answer 1

The correct option is B $1.8\times {10}^{-5}$

As the concentration of $H^{+}$ from acids is $0.2\times 0.95\times {10}^{-2}=1.9\times {10}^{-3}$ M hence $H^{+}$ from self-ionization of water can be neglected.
$K_a=C{\alpha }^2$
$=0.2\times (0.0095{)}^2$
$=1.8\times {10}^{-5}$
​$C=0.2mole/lit$
or
$0.2moleli{t}^{-1}$