Question

1.0 g of a weak monobasic acid ($HA$), when dissolved in 150 mL water, lowers the freezing point by ${0.186}^{o}C$. Also 1.0 g of the same acid required 125 mL of a 0.10N Nate solution for complete neutralisation. Determine dissociation constant $\left(k_a\right)$ of the weak acid. $k_f$ of water is 1.86 K kg $mo{l}^{-1}$.

• A. $4.16\times {10}^{-3}$
• B. $2.38\times {10}^{-6}$
• C. $6.43\times {10}^{-3}$
• D. $3.78\times {10}^{-6}$

Answer 1

Answer: (A)

$4.16\times {10}^{-3}$

From equation reaction,

$meq.$ of acid $=125\times 0.1=12.5$

$Eq.wt.$of acid $=\frac{1000}{12.5}=80$= mol. wt.
Molarity $=\frac{1}{8}\times \frac{100}{15}=0.083$

$-\Delta T_t=i{k}_{t}m=(1+\alpha )\times 1.86\times 0.083$

$\alpha =0.2$

$K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.083\times (0.2{)}^2}{0.8}=4.16\times {10}^{-3}$

Hence, the correct option is $\text{A}$