1.3245 g of a monobasic acid when dissolved in 100 g of water lowers the freezing point by ${0.2046}^{\circ }C$ , when solution containing 0.2 g of same acid is titrated required $\mathbf{15}.\mathbf{1}\mathbf{ml}$ of $\frac{\text{N}}{\text{10}}$  alkali (Assuming molarity = molality), the pH of the 1.3245 g of monobasic acid solution is: ($k_f$ of water $\mathbf{1}.\mathbf{86}\mathbf{K}\mathbf{kg}{\text{\hspace{0.17em}mol}}^{\text{-1}}$ )

Meq. of alkali = Meq. of acid
 $\Rightarrow 15.1\times \frac{1}{10}=\frac{0.2}{M}\times 1000$
On, solving M = 132.45 g${\text{mol}}^{\text{-1}}$ 
Now, molality of acid solution $\left(m\right)=\frac{1.3245\times 1000}{132.45\times 100}=0.1$
$\therefore {\text{\hspace{0.17em}\hspace{0.17em}ΔT}}_{\text{f}}{\text{=i×K}}_{\text{f}}\text{×m}$

$$\begin{gathered} 0.2048=i\times 1.86\times 0.1 \\ I=1.1=1+\alpha \\ \therefore \alpha =0.1 \end{gathered}$$

$\begin{array}{l} HA\rightleftharpoons H^{+}+ A^{-}\\ Now,C0 0\\ C-C\alpha C-C\alpha C\alpha \end{array}$

$\left[H^{+}\right]=C\alpha =0.1\times 0.1={10}^{-2} \therefore pH=2$