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REPEATER 2024 — Chemistry STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryREPEATER 20243 Marks
Question
Derive Ostwald's dilution law for weak acid. Obtain relation between solubility product and its solubility for $Al ( OH )_3$.
✓

Answer

Derivation of Ostwald's dilution law for weak acid :
Consider an equilibrium of weak acid $H A$ that exists in solution partly as the undissociated species $H A$ and partly $H^{\oplus}$ and $A^{\ominus}$ ions. Then,
$H A_{(a q)} \rightleftharpoons H_{(a q)}^{\oplus}+A_{(a q)}^{\ominus}$
$\therefore $ The acid dissociation constant is given by
$K_a=\frac{\left[H^{\oplus}\right]\left[A^{\ominus}\right]}{[H A]} \ldots\ldots(1)$
Suppose 1 mol of acid $H A$ is initially present in volume $V d m^3$ of the solution. At equilibrium the fraction dissociated would be $\propto$, where $\propto$ is degree of dissociation of the acid. The fraction of an acid that remains undissociated would be $(1\ - \propto$).
If $c$ is the initial concentration of an acid in $mol \ dm ^{-3}$ and $V$ is the volume in $dm ^3 mol^{-1}$ then $c=\frac{1}{V}$.
HA$H^{\oplus}$$A^{\ominus}$
Amount present at equilibrium/mol$1-\propto$$\propto$$\propto$
concentration at equilibrium/mol dm³$\frac{1-\alpha}{v}$ or $(1-\alpha) c$$\frac{\alpha}{v}$ or $\propto c$$\frac{\alpha}{v}$ or $\propto c$
Substituting the equilibrium concentration in equation (1), we get
$K_a=\frac{\alpha_c. \alpha c}{(1-\alpha) c}$
$\therefore \ K_a=\frac{\alpha^2 c}{(1-\alpha)}$
But for weak acid $H A, \propto$ is very small, $\quad \therefore 1-\propto \cong 1$
$\begin{array}{ll}\therefore & K_a=\alpha^2 c \\
\therefore & \alpha=\sqrt{\frac{K_a}{c}} \ \text { or } \ \propto=\sqrt{K_a \cdot V}\end{array}$
This is the required expression of Ostwald's dilution law for weak acid.
Relationship between solubility and solubility product for $Al ( OH )_3$ :
Solubility equilibrium for $Al ( OH )_3$ is given by,
$\begin{array}{ll} & Al(OH)_{3_{(s)}} \rightleftharpoons Al_{(a q)}^{3 \oplus_{(a q)}}+3 OH_{(a q)}^{\ominus} \\
\therefore & x=1, \ y=3 \\
\therefore & K_{s p}=(1)^1(3)^3 S^{1+3} \\
\therefore & K_{s p}=27 S^4\end{array}$
Above expression shows the relationship between solubility and solubility product for $Al ( OH )_3$.

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