Question

Nicotinic acid (Ka = 1.4 x ${10}^{-5}$) is represented by the formula HNic. For a solution which contains 0.10 mol of nicotinic acid per 2.0 litre of solution, the percent dissociation is:

• A. 4.2%
• B. 12.77%
• C. 1.66%
• D. 3.33%

Answer 1

The correct option is C

1.66%

Let the initial concentration of Nicotinic acid be C molL^-1
Let us write down the equation:

HNic $\leftrightharpoons$ ${Nic}^{-}$ + $H^{+}$

(1- $\alpha$)C $\alpha C$ $\alpha C$

$Ka=\frac{\left[C\alpha \right]\left[C\alpha \right]}{C(1-\alpha )}$
$1-\alpha =1$
$Ka=C{\alpha }^2$
$=1.4\times {10}^{-5}=0.05\times {\alpha }^2$
${\alpha }^2=\frac{1.4\times {10}^{-5}}{0.05}$
$=0.28\times {10}^{-3}$
$\alpha =0.01673$

Percentage dissociation=1.67 %