Ostwald Dilution Law was discovered by the German scientist Wilhelm Ostwald in the year of 1896. This law describes the degree of dissociation and its dependence on the concentration of the weak electrolytes. Basically Ostwald discovered this theory for the weak electrolytes to know their nature and dissociation and how they ionize which is very helpful in the advancement of further discoveries of acid-base chemistry and the Solutions. Before the Ostwald discovered T this law the nature of weak electrolytes was not known. Ostwald provides a pathway for under the concept of weak electrolytes and the way They ionize in the dilute or the concentrated Solutions.
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The pH of Weak Acids
Weak acids are those acids that dissociate partially in solutions. For example:
$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COOH}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\frac{\mathrm{c} \alpha \cdot \mathrm{c} \alpha}{\mathrm{c}(1-\alpha)}=\frac{\mathrm{c} \alpha^2}{1-\alpha} \\
& \mathrm{K}_{\mathrm{a}}=\mathrm{c} \alpha^2 \quad(\text { as }(1-\alpha) \approx 1)
\end{aligned}
$
Now, as we have given :
$
\begin{aligned}
& \mathrm{c}=0.002 \mathrm{~N} \text { or } 0.002 \mathrm{M} \quad(\text { Normality }=\text { Molarity, as } \mathrm{n} \text { factor }=1) \\
& \alpha=\frac{2}{100}=0.02
\end{aligned}
$
Thus, $\left[\mathrm{H}^{+}\right]=0.002 \times 0.02=4 \times 10^{-5}$
$
\begin{aligned}
& \mathrm{pH}=-\log _{10}\left(4 \times 10^{-5}\right) \\
& \mathrm{pH}=5-\log 4=4.4
\end{aligned}
$
Ostwald's Dilution Law
This is an application of the law of mass action for weak electrolyte dissociation equilibria. Consider ionization of a weak electrolyte say a monoprotic acid, acid HA.$\mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})$
Thus,
$\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}$
Moles before dissociation 1 0 0
Moles after dissociation 1 - ? ? ?
is the degree of dissociation of weak acid HA and c is the concentration.
Thus, according to the equilibrium constant equation, we have:
$\begin{aligned} & \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{\mathrm{c} \alpha \cdot \mathrm{c} \alpha}{\mathrm{c}(1-\alpha)} \\ & \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{c} \alpha^2}{(1-\alpha)}\end{aligned}$
For weak electrolytes, ? is small, thus 1 -? = 1$\mathrm{K}_{\mathrm{a}}=\mathrm{c} \alpha^2$ or $\alpha=\sqrt{\left(\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}\right)}$
A similar expression can be made for a weak base as BOH:$\mathrm{BOH}(\mathrm{aq}) \rightleftharpoons \mathrm{B}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}$
Similarly, for the base BOH, the expression of Kb can be written as
$
\begin{aligned}
& \mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \\
& \mathrm{K}_{\mathrm{b}}=\frac{\mathrm{c}^2}{(1-\alpha)}
\end{aligned}
$
Thus, if $1-$? $=1$, then
$
\mathrm{K}_{\mathrm{b}}=\mathrm{c} \alpha^2 \text { or } \alpha=\sqrt{\left(\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{c}}\right)}
$
Thus, if 1 -? = 1, then $\mathrm{K}_{\mathrm{b}}=\mathrm{c} \alpha^2$ or $\alpha=\sqrt{\left(\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{c}}\right)}$
From the expression for Ka or Kb it is evident that
(1) As the value of concentration decreases, the degree of dissociation increases
(2) As the value of concentration increases, the degree of dissociation decreases
This is called Ostwald's dilution law for weak electrolytes
Example.1
1. Calculate the amount of acetic acid in one liter of its solution. Given, the degree of dissociation is 2% and $\mathrm{Ka}=1.8 \times 10^{-5}$
1) (correct)2.7 g
2)27 g
3)10.8 g
4)5.4 g
Solution
According to the question,
$
\begin{array}{ccc}
\\
\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} \\
1 & 0 & 0 \\
1-\alpha & \alpha & \alpha
\end{array}
$
nere $\alpha$ is the degree of dissociation.
ow,
$
\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\frac{\mathrm{c} \alpha \cdot \mathrm{c} \alpha}{\mathrm{c}(1-\alpha)}=\frac{\mathrm{c} \alpha^2}{(1-\alpha)}
$
As the dissociation constant is very less than 1 , so $1-\alpha \approx 1$
$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=\mathrm{c} \alpha^2 \\
& \Rightarrow 1.8 \times 10^{-5}=\mathrm{c} \times\left(\frac{2}{100}\right)^2 \\
& =0.045 \mathrm{~mole} / \text { litre }
\end{aligned}
$
Amount of acetic acid in 1 litre $=60 \times 0.045=2.7 \mathrm{~g}$
Hence, the answer is the option (1).
Example.2
2. A weak acid HA has $\mathrm{Ka}=1.4 \times 10^{-5}$. Calculate its percent dissociation in a solution that contains 0.1 moles of HA per 2 liters of solution.
1)2.8%
2) (correct)1.67%
3)28%
4)Data insufficient
Solution
Given,
$
[\mathrm{HA}]=\frac{0.1}{2}=0.05 \mathrm{M}
$
For weak acid HA
$
\begin{aligned}
& \mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-} \\
& \begin{array}{ll}
\text { c } \quad 0 & 0
\end{array} \\
& \mathrm{c}(1-\alpha) \quad \mathrm{c} \alpha \quad \mathrm{c} \alpha \\
& \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{c} \alpha \cdot \mathrm{c} \alpha}{\mathrm{c}(1-\alpha)}=\frac{\mathrm{c} \alpha^2}{(1-\alpha)} \\
&
\end{aligned}
$
As the dissociation constant is very less than 1 , so $1-\alpha \approx 1$
$
\begin{aligned}
& \therefore \mathrm{K}_{\mathrm{a}}=\mathrm{c} \alpha^2 \\
& \Rightarrow 1.4 \times 10^{-5}=0.05 \times \alpha^2 \\
& \Rightarrow \alpha=1.67 \times 10^{-2}
\end{aligned}
$
The degree of dissociation in % will be 1.67.
Hence, the answer is the option (2).
Example.3
3. Calculate the concentration of fluoroacetic acid when $\left[\mathrm{H}^{+}\right]=2 \times 10^{-3} \mathrm{M}$.
Given, Ka of acid $=2.8 \times 10^{-3}$
1) $2 \times 10^{-3} \mathrm{M}$
2) $2.8 \times 10^{-3} \mathrm{M}$
3) (correct) $3.4 \times 10^{-3} \mathrm{M}$
4) $4.8 \times 10^{-5} \mathrm{M}$
Solution
Given:
$
\left[\mathrm{H}^{+}\right]=2 \times 10^{-3} \mathrm{M}
$
Ka of acid $=2.8 \times 10^{-3}$
For Fluoroacetic acid:
$
\mathrm{CH}_2 \mathrm{FCOOH} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CH}_2 \mathrm{FCOO}^{-}
$
$c$ 0 0
$c(1-\alpha)$ $c \alpha$ $c \alpha$
From the above reaction, it is clear that
$
\left[\mathrm{H}^{+}\right]=\mathrm{c} \alpha=2 \times 10^{-3}
$
Now,
$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{c} \alpha \cdot \mathrm{c} \alpha}{\mathrm{c}(1-\alpha)}=\frac{\mathrm{c} \alpha^2}{(1-\alpha)} \\
& 2.8 \times 10^{-3}=\frac{2 \times 10^{-3} \alpha}{(1-\alpha)} \longrightarrow(i)
\end{aligned}
$
Here we can see $\alpha$ is not small and hence it cannot be neglected in comparison to 1.
Solving equation (i) we get,
$
\alpha=0.583
$
Now, it is given that
$
\begin{aligned}
& {\left[\mathrm{H}^{+}\right]=\mathrm{c} \alpha=2 \times 10^{-3}} \\
& \Rightarrow c=\frac{2 \times 10^{-3}}{0.583} \\
& \Rightarrow \mathrm{c}=3.43 \times 10^{-3} \mathrm{M}
\end{aligned}
$
Hence, the answer is the option (3).
Example.4
4. What is true regarding Ostwald's dilution law:
1) The value of Ka tends to zero for strong electrolytes
2)The smaller the value of Ka, the stronger the acid
3)This law is valid for strong as well as weak electrolytes
4) (correct)This law is only valid for weak electrolytes
Solution
Ostwald's dilution law is only valid for weak electrolytes.
The value of Ka is high for strong acids.
Hence, the answer is the option(4).
5. The first and second dissociation constants of an acid $H_2 A$ are $1.0 \times 10^{-5}$ and $5.0 \times 10^{-10}$ respectively. The overall dissociation constant of the acid will be
1) $0.2 \times 10^5$
2) $5.0 \times 10^{-5}$
3) $5.0 \times 10^{15}$
4) $\left(\right.$ correct) $5.0 \times 10^{-15}$
Solution
As we have learned in concept:
pH of Solutions: Weak Acids -
the pH of Weak Acids
Weak acids are those acids that dissociate partially in solutions. For example:
$
\begin{aligned}
& \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COOH}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\frac{\mathrm{c} \alpha \cdot \mathrm{c} \alpha}{\mathrm{c}(1-\alpha)}=\frac{\mathrm{c} \alpha^2}{1-\alpha} \\
& \mathrm{K}_{\mathrm{a}}=\mathrm{c} \alpha^2 \quad(\mathrm{as}(1-\alpha) \approx 1)
\end{aligned}
$
Now, as we have given :
$
\begin{aligned}
& \mathrm{c}=0.002 \mathrm{~N} \text { or } 0.002 \mathrm{M} \quad(\text { Normality }=\text { Molarity, as } \mathrm{n} \text { factor }=1) \\
& \alpha=\frac{2}{100}=0.02
\end{aligned}
$
$
\begin{aligned}
& \text { Thus, }\left[\mathrm{H}^{+}\right]=0.002 \times 0.02=4 \times 10^{-5} \\
& \mathrm{pH}=-\log _{10}\left(4 \times 10^{-5}\right) \\
& \mathrm{pH}=5-\log 4=4.4
\end{aligned}
$
$\begin{aligned} & H_2 A \rightleftharpoons H^{+}+H A^{-}; \\ & K_1=\frac{\left[H^{+}\right]\left[H A^{-}\right]}{\left[H_2 A\right]}=1 \times 10^{-5} \\ & H A \rightleftharpoons H^{+}+A^{2-}; K_2=5 \times 10^{-10}=\frac{\left[H^{+}\right]\left[A^{2-}\right]}{\left[H A^{-}\right]} \\ & K=\frac{\left[H^{+}\right]^2\left[A^{2-}\right]}{\left[H_2 A\right]}=K_1 \times K_2=1 \times 10^{-5} \times 5 \times 10^{-10} \\ & =5 \times 10^{-15}\end{aligned}$
Hence, the answer is the option (4).
Ostwald's Dilution Law is basically the study of weak electrolytes in which we study their ionization and the behavior of weak electrolytes only, which helps to control and prediction of chemical and various biological processes. The ionization of weak electrolytes is characterized by an equilibrium between the ionized and un-ionized forms. Understanding the ionization of weak acids and bases is crucial in preparing buffer solutions that resist changes in pH. In pharmaceuticals, the ionization of weak acids and bases affects drug absorption and distribution in the body. Many industrial processes rely on controlling the ionization of weak electrolytes for optimal reactions and product yields.
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