Question

Find the dissociation constant $K_a$ of a weak monobasic acid which is 3.5 % dissociated in a $\frac{N}{10}$ solution at ${25}^{\circ }C.$

• A. $1.23\times {10}^{-4}$
• B. $1.26\times {10}^{-6}$
• C. $3.40\times {10}^{-4}$
• D. $4.20\times {10}^{-3}$

Answer 1

The correct option is A $1.23\times {10}^{-4}$

$C=\frac{1}{10}=0.1M$
percentage dissociation = 3.5 %
degree of dissociation$\left(\alpha \right)=\frac{3.5}{100}=0.035$
we know the relation for weak acid, $\alpha =\sqrt{\frac{K_a}{C}}$
${\alpha }^2=\frac{K_a}{C}$
$K_a={\alpha }^2\times C$
$K_a=(0.035{)}^2\times 0.1$
$K_a=0.001225\times 0.1$
$K_a=0.0001225$
$K_a=1.225\times {10}^{-4}$