Question

$2\mathrm{molal}$ solution of a weak acid $\mathrm{HA}$ has a freezing point of $3.885{}^{\mathrm{o}}\mathrm{C}$. The degree of dissociation of this acid is $\mathrm{\_\_\_\_\_\_\_\_\_}\times {10}^{–3}$. (Round off to the Nearest Integer). [Given: Molal depression constant of water $=1.85\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{–1}$, Freezing point of pure water $=0{}^{\mathrm{o}}\mathrm{C}$]

Answer 1

Step 1: Given data:

Molality of the weak acid solution is $\mathrm{m}=2\mathrm{molal}$

Depression in freezing point is $∆{\mathrm{T}}_{\mathrm{f}}=3.885{}^{\mathrm{o}}\mathrm{C}$

$∆{\mathrm{T}}_{\mathrm{f}}=3.885{}^{\mathrm{o}}\mathrm{C}$

Molal depression constant of water is ${\mathrm{k}}_{\mathrm{f}}=1.85\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}$

Freezing point of pure water is ${\mathrm{T}}_{\mathrm{f}}=0{}^{\mathrm{o}}\mathrm{C}$

Step 2: Find the van't Hoff factor $\left(i\right)$:

$∆{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{k}}_{\mathrm{f}}\times \mathrm{m}$

Where, $∆{\mathrm{T}}_{\mathrm{f}}$ is the Depression in freezing point

$\mathrm{i}$ is the van't Hoff factor

${\mathrm{k}}_{\mathrm{f}}$ is the molal depression constant

$\mathrm{m}$ is the molality of the solution

$$\begin{gathered} \Rightarrow 3.885°\mathrm{C}=\mathrm{i}\times 1.85\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}\times 2\mathrm{molal}\left[\because 1\mathrm{molal}=1\mathrm{kg}{\mathrm{mol}}^{-1}\right] \\ \Rightarrow \mathrm{i}=\frac{3.885}{1.85\times 2} \\ \Rightarrow \mathrm{i}=1.05 \end{gathered}$$

Step 3: Find the degree of dissociation $\left(\mathit{\alpha }\right)$:

$i\mathit{}\mathit{=}\mathit{}\mathit{1}\mathit{}\mathit{+}\mathit{}\left(n-1\right)\mathit{}\alpha$

Where, $\mathrm{i}$ is the van't Hoff factor

$\mathrm{\alpha }$ is the degree of dissociation

$\mathrm{n}$ is the number of ions dissociated from the weak acid

i.e., $\mathrm{HA}\to {\mathrm{H}}^{+}+{\mathrm{A}}^{-}$

Here, the number of ions after the dissociation is $2$.

Hence, $\mathrm{n}=2$

By substituting the above-given values in the equation;

$$\begin{gathered} \mathit{\Rightarrow }\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{1}\mathit{.}\mathit{05}\mathit{}\mathit{=}\mathit{}\mathit{1}\mathit{}\mathit{+}\mathit{}\mathit{(}\mathit{2}\mathit{-}\mathit{1}\mathit{)}\mathit{}\alpha \\ \mathit{\Rightarrow }\mathit{(}\mathit{2}\mathit{-}\mathit{1}\mathit{)}\mathit{}\alpha \mathit{}\mathit{=}\mathit{}\mathit{1}\mathit{.}\mathit{05}\mathit{}\mathit{-}\mathit{1} \\ \mathit{\Rightarrow }\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\alpha \mathit{}\mathit{=}\mathit{}\mathit{0}\mathit{.}\mathit{05} \\ \mathit{\Rightarrow }\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\alpha \mathit{}\mathit{=}\mathit{}\mathit{50}\mathit{}\mathit{\times }\mathit{}{\mathit{10}}^{\mathit{-}\mathit{3}} \end{gathered}$$

Hence, the degree of dissociation of $\mathbf{2}\mathbf{}\mathbf{molal}$ solution of a weak acid $\mathbf{HA}$ having a freezing point of $\mathbf{3}\mathbf{.}\mathbf{885}\mathbf{}{}^{\mathbf{o}}\mathbf{C}$ is $\mathit{\alpha }\mathit{}\mathit{=}\mathit{}\underline{\mathit{50}}\mathit{}\mathit{\times }\mathit{}{\mathit{10}}^{\mathbf{-}\mathbf{3}}$.