chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7-8-xcpnylhmrt0

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Salt of weak acid and strong base </primary>
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- $[OH^-] = \sqrt {K_h \times C_{salt}}$
- $K_h = \frac {K_w}{K_a}$
- $[OH^-]=\sqrt{\frac {K_w} {K_a} \times C_{salt}}$
- $log [OH^-]=\frac{1}{2}[log k_w-log k_a+log C_{salt}$
- $-log [OH^-]=\frac{1}{2}[-log k_w+log k_a-log C_{salt}$
- $pOH=\frac{1}{2}[pK_W-pK_a-log C_{salt}$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-10.png)
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<footer style = "font-size: 18px">Weak electrolytes $\rightarrow$ Recap $\rightarrow$ <span style = "color:blue"> Salt of weak acid and strong base </span> $\rightarrow$ pH of salt of strong acid and weak base $(NH_4 Cl)$ $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> pH of salt of strong acid and weak base $(NH_4 Cl)$ </primary>
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- $[H^{+}] =\sqrt {K_h \times C_{salt}}$ 
- $[H^{+}] =\frac {K_w}{K_b} \times C_{salt}$
- $-log [H^+]=-\frac{1}{2} log K_\omega+\frac{1}{2} log K_b-\frac{1}{2}logC_{salt}$
- $p H=\frac{1}{2} pK_W-\frac{1}{2} pK_b-\frac{1}{2} log C_{salt}$

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<footer style = "font-size: 18px">Recap $\rightarrow$ Salt of weak acid and strong base $\rightarrow$ <span style = "color:blue"> pH of salt of strong acid and weak base $(NH_4 Cl)$ </span> $\rightarrow$ Problem with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **Question** What is the pH of a 0.5 M aqueous NaCN solution
- $pk_b$ of $C N^-$is 4.70 
- **Answer**
- $[NaCN]  =0.5 M$
- $[salt]  =0.5 M$
- NaCN is salt of strong base and weak acid that is HCN
- $[OH^-]=\sqrt {K_h \times C_{salt}}$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-12.png)

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<footer style = "font-size: 18px">Salt of weak acid and strong base $\rightarrow$ pH of salt of strong acid and weak base $(NH_4 Cl)$ $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Problem with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **Answer**
- $[OH^-]=\sqrt {K_h \times C_{salt}}$
- $-log [OH^-]=-\frac{1}{2} \log K_h-\frac{1}{2} log C_{salt}$
- $pOH=\frac{1}{2} p K_h-\frac{1}{2} log C_{salt}$
- $p k_h=p k_b  of  CN^-= 4.70$
- $pOH=\frac{1}{2}(4.70)-\frac{1}{2} log (0.5)$
- $pOH \longrightarrow p^H $
- $p^H=14-pOH$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-14.png)

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<footer style = "font-size: 18px">Problem with solution $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Problem with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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<main>

- **Question** The dissociation constant of a substituted benzoic acid at $25^{\circ} \mathrm{C}$ is $1\times 10^{-4}$ The pH of a 0.01 M solution of its sodium salt is
- **Answer**
- Sodium salt of weak acid
- $K_a=1 \times 10^{-4}$
- $C_{acid}$ =0.01 M 
- To find pH of the solution

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-15.png)

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- Sodium salt of weak acid 
- $[OH^-]=\sqrt{K_h \times C_{salt}}$
- $[OH^-]=\sqrt{\frac{K_W}{K_a} \times 0.01}$
- $[OH^-]=\sqrt{\frac{1 \times 10^{-14}}{1 \times 10^{-4}} \times 10^{-2}}$
- $[OH^-]=\sqrt{10^{-12}}=10^{-6}$
- $[OH^{-}]=10^{-6} $
- pOH=6 
- pH=8

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-16.png)

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### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **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

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-17.png)

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<footer style = "font-size: 18px">Problem with solution $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Titration of a weak acid with a strong base $\rightarrow$ Buffer solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Titration </primary>
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- $CH_3 COONa \rightarrow CH_3COO^- + H^+$

- $CH_3COO^- = C_{salt}$

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

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

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

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

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

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<footer style = "font-size: 18px">Buffer solution $\rightarrow$ Titration $\rightarrow$ <span style = "color:blue"> Titration </span> $\rightarrow$ Titration $\rightarrow$ Titration </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Titration </primary>
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- $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$

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

- $\simeq C_{acid}$

- $\alpha < < < 1$

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

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

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

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-22.png)

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<footer style = "font-size: 18px">Titration $\rightarrow$ Titration $\rightarrow$ <span style = "color:blue"> Titration </span> $\rightarrow$ Titration $\rightarrow$ Titration of weak acid with a strong base </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Titration </primary>
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<main>

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

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

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

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

- Handerson Hassalbach equation for buffer

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-23.png)

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<footer style = "font-size: 18px">Titration $\rightarrow$ Titration $\rightarrow$ <span style = "color:blue"> Titration </span> $\rightarrow$ Titration of weak acid with a strong base $\rightarrow$ Titration of weak acid with a strong base </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Titration of weak acid with a strong base </primary>
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<main>

- 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

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-26.png)

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<footer style = "font-size: 18px">Titration of weak acid with a strong base $\rightarrow$ Titration of weak acid with a strong base $\rightarrow$ <span style = "color:blue"> Titration of weak acid with a strong base </span> $\rightarrow$ Titration of weak acid with a strong base $\rightarrow$ Titration of weak acid with a strong base </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Titration of weak acid with a strong base </primary>
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- 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

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-27.png)

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### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Titration of weak acid with a strong base </primary>
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- In the solution f 4 mmol of acetic acid and 1 mmol of sodium acetate

- By Handerson equation

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

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

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

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

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

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-28.png)

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### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Titration of weak acid with a strong base </primary>
<hr>
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- 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

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

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

- $pH =4.73+\log 1$

- $pH =4.73$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-29.png)

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### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Titration of weak acid with a strong base </primary>
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- 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 $CH_3 COONa$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-30.png)

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### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Titration of weak acid with a strong base </primary>
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<main>

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

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

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

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

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

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

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

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-31.png)

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### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Titration of weak acid with a strong base </primary>
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- 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

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-32.png)

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### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **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

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-34.png)

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<footer style = "font-size: 18px">Titration of weak acid with a strong base $\rightarrow$ Titration of weak acid with a strong base $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Problem with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
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- **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)

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

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-35.png)

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<footer style = "font-size: 18px">Titration of weak acid with a strong base $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Problem with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-7 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

- **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

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-07-chapter-07-ionic-equillibrium-l-7_8-xcpnylhmrt0-36.png)

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