Ionization Of Acids And Bases

Ionization Of Weak Acids

The pH of Weak Acids
Weak acids are those acids that dissociate partially in solutions. For example:

• 8 M HA (K_a = 2 x 10^-8)
  The chemical equation for the dissociation of weak acid HA is as follows:
  
  Initial: 8M 0 0
  Equil: 8 - 8? 8? 8?
  
  The equilibrium constant K_a for the weak acid is given as follows:
  $\begin{aligned} & \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{8 \alpha \cdot 8 \alpha}{8(1-\alpha)}=\frac{8 \alpha^2}{1-\alpha} \\ & \mathrm{K}_{\mathrm{a}}=8 \alpha^2 \quad(\text { as }(1-\alpha) \approx 1) \\ & \text { Thus, } \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{8}}=\sqrt{\frac{2 \times 10^{-8}}{8}}=\sqrt{\frac{10^{-8}}{4}}=0.5 \times 10^{-4} \\ & {\left[\mathrm{H}^{+}\right]=8 \times 0.5 \times 10^{-4}=4 \times 10^{-4}} \\ & \mathrm{pH}=-\log _{10} 4+4 \\ & \mathrm{pH}=-0.60+4=3.4\end{aligned}$
  Thus, the pH of this given acid = 3.4
• 0.002N CH₃COOH(? = 0.02)
  The chemical equation for the dissociation of CH₃COOH is as follows:
  $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$
  Initial: c 0 0
  Equil: c - c? c? c?
  
  The equilibrium constant K_a for the weak acid is given as follows:

  $
  \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}
  $

  Thus, the pH of acetic acid = 4.4

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 K_b 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 K_b 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

Recommended topic video on ( Ionization of Acids and Bases)

Some Solved Examples

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 K_a tends to zero for strong electrolytes

2)The smaller the value of K_a, 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 K_a 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:

• 8 M HA (K_a =2 x 10^-8)
  The chemical equation for the dissociation of weak acid HA is as follows:
  $\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}$
  Initial: 8M 0 0
  Equil: 8 - 8? 8? 8?
  
  The equilibrium constant K_a for the weak acid is given as follows:
  $\begin{aligned} & \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{8 \alpha .8 \alpha}{8(1-\alpha)}=\frac{8 \alpha^2}{1-\alpha} \\ & \mathrm{K}_{\mathrm{a}}=8 \alpha^2 \quad(\text { as }(1-\alpha) \approx 1) \\ & \text { Thus, } \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{8}}=\sqrt{\frac{2 \times 10^{-8}}{8}}=\sqrt{\frac{10^{-8}}{4}}=0.5 \times 10^{-4} \\ & {\left[\mathrm{H}^{+}\right]=8 \times 0.5 \times 10^{-4}=4 \times 10^{-4}} \\ & \mathrm{pH}=-\log _{10} 4+4 \\ & \mathrm{pH}=-0.60+4=3.4\end{aligned}$
  Thus, the pH of this given acid = 3.4
• 0.002N CH₃COOH(? = 0.02)
  The chemical equation for the dissociation of CH₃COOH is as follows:
  $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$
  Initial: c 0 0
  Equil: c - c? c? c?
  
  The equilibrium constant K_a for the weak acid is given as follows:

  $
  \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}
  $

  Thus, the pH of acetic acid = 4.4

$\begin{aligned} & H_2 A \rightleftharpoons H^{+}+H A^{-}; \\ & K_1=\frac{\left[H^{+}\right]\left[H A^{-}\right]}{\left[H_2 A\right]}=1 \times 10^{-5} \\ & H A \rightleftharpoons H^{+}+A^{2-}; K_2=5 \times 10^{-10}=\frac{\left[H^{+}\right]\left[A^{2-}\right]}{\left[H A^{-}\right]} \\ & K=\frac{\left[H^{+}\right]^2\left[A^{2-}\right]}{\left[H_2 A\right]}=K_1 \times K_2=1 \times 10^{-5} \times 5 \times 10^{-10} \\ & =5 \times 10^{-15}\end{aligned}$

Hence, the answer is the option (4).

Summary

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.