Concept of ionic equilibrium comes when we are dealing with weak electrolytes

Weak electrolytes

• Weak bases
• Weak acid
• Salt of weak acid and weak bases

Weak acids

• $HA \rightleftharpoons H^+ + A^-$

Weak bases

• $BOH \rightleftharpoons B^+ + OH^-$

Salt of weak acid

• $NaA \rightarrow Na^+ + A^-$

• $\underbrace{A^- + H_2O \rightleftharpoons HA + OH^-}_{Equilibrium}$

Weak acid

• $[H^+]=\sqrt{K_a \times C_{HA}}$

Weak base

• $[OH^-] = \sqrt {K_b \times C_{BOH}}$

Question The pH of 0.1M solution of the acid HQ is 3 The value of ionization constant, $K_a$ of the acid is

Answer

$H^{+}=\sqrt{K_a \times C_{a c i d}}$

$H^{+}=\sqrt{K_a \times 0.1}$

pH=3

$[H^+]=1 \times 10^{-3}$

$pH=-log [H^{+}]$

$1 \times 10^{-3}=\sqrt{K_a \times 0.1}$

$K_a \times 0.1=1 \times 10^{-6}$

$K_a=1 \times 10^{-6} / 0.1$

$K_a=1 \times 10^{-5}$ M

Buffer solution

Used to maintain pH of a solution

Small addition of acid or base does not affect the pH of the solution

Weak acid in presence of salt of weak acid with strong base

$CH_3 COOH + CH_3COONa$

$\downarrow\quad\quad\quad\quad\quad\quad\quad\downarrow$

Weak acid$\quad\quad$ Salt of weak acid with a strong base

$CH_3 COONa \rightarrow CH_3COO^- + H^+$

$CH_3COO^- = C_{salt}$

$CH_3COOH \rightleftharpoons CH_2COO^{-} + H^+$

$C_{acid} \quad\quad\quad\quad \quad 0 \quad\quad\quad\quad \quad 0$

$C_{acid}(1-\alpha)\quad\quad C\alpha \quad\quad\quad C\alpha$

$[CH_3COO^-] \simeq C\alpha+C_{salt}$

$[CH_3COO^-]\simeq C_{salt}$

$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$

$C(1-\alpha) \quad\quad\quad C_{\text {salt }} \quad \quad \quad C \alpha$ $-$ equilibrium

$\simeq C_{acid}$

$\alpha < < < 1$

$C(1-\alpha)\simeq C_{acid}$

$K_a=\frac{[CH_3COO^{-}][H^{+}]}{[CH_3COOH]}$

$K_a=\frac{C_{\text {salt }}\left[H^{+}\right]} {C_{acid}}$

$K_a=\frac{C_{\text {salt }}\left[\mathrm{H}^{+}\right]}{C_{\text {acid }}}$

$[H^+]=\frac {K_a \times Cacid}{C_{{salt }}}$

$-\log [H^+]=-\log K_a-\log [ acid ]$ $+\log [ salt ]$

$p H=p k_a+\log \frac{[{ salt}]}{[a c i d]}$

Handerson Hassalbach equation for buffer

$CH_3 COOH + NaOH$

What will happen when we add NaOH from time to time

What is the change in pH

50 ml of 0.1 M $CH_3 COOH$ is titrated with 50 ml of 0.1 M NaOH

$K_a$ for acetic acid = 1.8 $\times 10^{-5} molL^{-1}$

Before the addition of NaOH, there is 50 ml of 0.1 M $CH_3 COOH$

$pK_a = - log 1.8 \times 10^{-5}$

$pK_a = 4.73$

$H^+=\sqrt{K_a \times C_{a c i d}}$

$-\log [H^{+}]=-\frac{1}{2} \log K_a-\frac{1}{2} \log [ acid ]$

$p H=\frac{1}{2} \log K_a-\frac{1}{2} \log (0.1)$

pH of the salt can be calculated

We started adding NaOH solution

Suppose we added 10 ml of 0.1 M NaOH

$CH_3 COOH+NaOH \rightarrow CH_3COONa+H_2 O$

$50 \times 0.1 \quad \quad 10 \times 0.1$ $\quad\quad \quad \quad\quad\quad\quad\quad\quad\quad$ - in mmol

5 mmol $\quad \quad$1 mmol

(5-1)=4 mmol $\quad$ 0 $\quad\quad\quad\quad\quad$ 1 mmol

In the solution f 4 mmol of acetic acid and 1 mmol of sodium acetate

By Handerson equation

$p H=p K_a+\log \frac{[salt]}{[{ acid }]}$

$\operatorname{conc}^n =\frac{{ mmol }}{{ volume\hspace{0.5mm} in \hspace{0.5mm}ml }}$

${conc}^n =\frac{1}{50+10}=\frac{1}{60}$

$p H=4.73+\log \frac{1 / 60}{4 / 60}$

$p H=4.73+\log 1 / 4=4.73-\log 4$

50 ml of 0.1 M $CH_3 COOH$ with 25 ml of 0.1 M NaOH

$CH_3 COOH+NaOH \longrightarrow CH_3COONa+H_2 O$

$5 mmol \quad \quad \quad2.5 mmol$

(5-2.5) $\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad$ 2.5 mmol

2.5 mmol

$pH=p K_a+\log \frac{[ { salt }]}{[{ acid }]}$

$pH=4.73+\log \frac{2.5 / 75}{2.5 / 75}$

$pH =4.73+\log 1$

$pH =4.73$

50 ml of 0.1 M $CH_3COOH$ with 50 ml of 0.1 M NaOH

• $CH_3 COOH+NaOH \longrightarrow CH_3COONa+H_2 O$

$5 mmol\quad\quad \quad 5mmol \quad\quad\quad\quad0\quad\quad\quad\quad\quad0$

$0 \quad\quad\quad\quad\quad\quad \quad 0 \quad\quad\quad\quad\quad\quad5mmol$

In a solution, we have 5 mmol of $CH_3 COONa$

5 mol of $CH_3 COONa$ in 100 ml of solution

Concentration of salt =$\frac{5}{100}=.05 M$

$[OH^{-}]=\sqrt{K_h \times C_{s a l t}}$

$[OH^{-}]=\sqrt{\frac{K_w}{K_a} \times C_{salt}}$

$pOH=\frac{1}{2} p K_W-\frac{1}{2} p K_a-\frac{1}{2} \log [{ salt }]$

$pOH=7-\frac{1}{2} \times 4.73-\frac{1}{2} \log [{salt }]$

$pOH \longrightarrow pH \quad pH=14-p^{OH}$

50 ml of 0.1 M $CH_3COOH$ with 60 ml of 0.1 M NaOH

$CH_3 COOH+NaOH \longrightarrow CH_3COONa+H_2O$

$5 mmol \quad\quad \quad\quad6 mmol\quad\quad0\quad\quad\quad\quad\quad\quad0$

$0 mmol \quad\quad \quad\quad1 mmol\quad\quad5 mmol\quad\quad\quad\quad0$

In the solution NaOH, $CH_3 COONa$ is present

NaOH + salt is of weak acid and strong base

$[OH^{-}]=C_{base}=\frac{1}{110}$ [V=50ml+60ml]

$-\log [OH^-]=pOH$

pH=14-pOH

Question 0.1 M NaOH is titrated with 0.1 M HA till the end point: Ka for HA is 5.6$\times$ $10^{-6}$ and degree of hydrolysis is less compared to 1. Calculate the pH of the resulting solution at the end point

Answer

$NaOH+HA \rightarrow NaA+H_2 O$

0.1M $\quad\quad$ 0.1M

xml $\quad\quad$ xml of HA

Answer

$NaOH+HA \rightarrow NaA+H_2O$

$\tiny Xml of 0.1 M NaOH$ $\tiny Xml of 0.1 M HA$

$0 \quad\quad\quad\quad 0 \quad \quad \quad \quad \frac {x \times 0.1}{mmol}$

In the solution, we have only salt (salt of weak acid and strong base)

• $[NaA]=\frac{x \times 0.1}{2 x}=\frac{0.1}{2}=.05 M$

Answer

Salt of weak acid and strong base

• $[OH^-]=\sqrt{{K_h} \times C_{\text {salt }}}$

$[OH^-]=\sqrt{\frac{K_W}{K_a} \times c_{s alt}}$

$[OH^-]=\sqrt{\frac{K_w}{K_a} \times 0.05}$

$pOH=\frac{1}{2} p K_w-\frac{1}{2} p K_a-\frac{1}{2} \log 0.05$

$-\log [OH]=-\frac{1}{2} \log K_w$ $+\frac{1}{2} \log K_a-\frac{1}{2} \log C_{salt}$

$pOH \rightarrow$ of the solution

Answer

Salt of weak acid and strong base

$[OH^{-}]=\sqrt {K_h \times C_{salt}}$

$[OH^{-}]= \sqrt{\frac{K_w}{K_a} \times C_{{salt }}}$