chemistry-class-11-unit-07-chapter-08-ionic-equillibrium-l-8-8-n3q5ut39bva

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

* 50 ml  of 0.11 $CH_3 COOH$ is titraqted with 0.1 M NaOH

* 0 ml of $NaOH  is added to the  weak acid solution

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

* 10 ml of $NaOH is added to the salt  $CH_3COONa+CH_3COOH$

- $CH_3COONa+CH_3COOH$ is buffer solution

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

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-03.jpg)

</div>

</main>
<hr>
<footer style = "font-size: 18px"> $\rightarrow$ Ionic equilibrium $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Concentration of solution $\rightarrow$ Titration </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Concentration of solution </primary>
<hr>
<main>

* At the equivalence point
* 50 ml of 0.1 M NaOH +50 ml of $0.1 \mathrm{M}$ acetic acid

- $Concentration =\frac{no.of mole of salt formed}{total volume}$

- $Concentration=\frac{5 {mmol of salt  }}{(100 {ml})} =.05 {mol}{L}$

- m mol=  Vol in  m l $\times$  molarity

- Molarity = m mol/vol in ml

</div>
	
<div style='float: right; width:0%;'>
	
![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-04.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Ionic equilibrium $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Concentration of solution </span> $\rightarrow$ Titration $\rightarrow$ Titration </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Solubility of a sparingly soluble salt </primary>
<hr>
<main>

- **Salts are of three types**

* Salts which are non-electrolyte

* Soluble salt $\xrightarrow {H_2O}$ breaks (complete dissociation)

- Example $NaCl \longrightarrow Na^{+}  (aq + Cl^{-} (aq)$

* Sparingly soluble salt - lot of salts are sparingly soluble

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-08.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Titration $\rightarrow$ Titration $\rightarrow$ <span style = "color:blue"> Solubility of a sparingly soluble salt </span> $\rightarrow$ Solubility of a sparingly soluble salt $\rightarrow$ Solubility of a sparingly soluble salt </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Solubility of a sparingly soluble salt </primary>
<hr>
<main>

* $AgCl \quad\quad\quad \rightleftharpoons \quad\quad\quad Ag^+(aq)+Cl^-(aq)$

- $\quad$ Undissociated $ \quad \quad \quad \quad \quad$  dissociated

* Undissociated and dissociated species are in equilibrium

* So, the concept of ionic equilibrium can be applied

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-10.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Titration $\rightarrow$ Solubility of a sparingly soluble salt $\rightarrow$ <span style = "color:blue"> Solubility of a sparingly soluble salt </span> $\rightarrow$ Solubility of a sparingly soluble salt $\rightarrow$ Solubility product </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Solubility of a sparingly soluble salt </primary>
<hr>
<main>

- $A g C l(s) \rightleftharpoons A g^{+}(a q)+Cl^{-}(a q)$

- Solubility of salt is S mol/l

- $AgCl(s) \rightleftharpoons Ag^{+}(a q)+Cl^{-}(a q)$

- $\quad  \quad \quad  \quad \quad   $ S mol/L of $Ag^{+}  \quad$ S mol/L of $ Cl^-$

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-11.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Solubility of a sparingly soluble salt $\rightarrow$ Solubility of a sparingly soluble salt $\rightarrow$ <span style = "color:blue"> Solubility of a sparingly soluble salt </span> $\rightarrow$ Solubility product $\rightarrow$ Solubility and solubility product </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Solubility product </primary>
<hr>
<main>

- $AgCl(s) \rightleftharpoons A g^{+}(aq)+Cl^{-}(aq)$

- $K=\frac{\left[\mathrm{Ag}^{\dagger}\right]\left[\mathrm{Cl}^{-}\right]}{\operatorname{AgCl}(\mathrm{s}) \sim 1}$

- $K_{sp}=[{Ag}^{+}][{Cl}^{-}]$

- $ K_{sp} $ is solubility product

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-12.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Solubility of a sparingly soluble salt $\rightarrow$ Solubility of a sparingly soluble salt $\rightarrow$ <span style = "color:blue"> Solubility product </span> $\rightarrow$ Solubility and solubility product $\rightarrow$ Solubility and solubility product </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

- **Problem** The solubility of $A_2 X_3$ is y mol $dm^{-3}$.Its solubility product is

- **Solution**

- $A_2X_3 \rightleftharpoons2 A^{3+}+3 X^{2-}$

- $\quad\quad\quad\quad $2y $\quad\quad\quad$ 3y

- $K_{s p}=[A^{3+}]^2[X^{2-}]^3$

- $K_{s p}=(2 y)^2(3 y)^3=4 y^2 \times 27 y^3$

- $K_{s p}=108 y^5$

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-15.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Solubility and solubility product $\rightarrow$ Solubility and solubility product $\rightarrow$ <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-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

- **Problem** For a sparingly soluble salt ApBq, the relationship of solubility product (Ls) and its solubility is

- **Solution**

- $A_p B_q \rightleftharpoons p A^{q +}+qB ^{P-}$

- $\quad\quad\quad\quad\quad\ pS\quad\quad qS$

- $Ls=[A^{q+}]^p[B^{p-}]^q$

- $Ls=(pS)^p(qS)^q$

- $Ls=p^p q^q S^{p+q}$

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-17.jpg)

</main>
<hr>
<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-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

- **Problem**  The molar solubility (in mol $L^{-1}$ )of a sparingly soluble salt $MX_4$ is ' s '. The corresponding solubility product is Ksp. 's' is given in terms of Ksp by the relation

- **Solution**

- $MX_4 \rightleftharpoons M^{+4}(a q)+4 X^{-}$
 
- $\quad \quad \quad  \quad \quad \quad S \quad \quad 4S $

- $K_{s p}=[M^{4+}] [X^{-}]^4$

- $K_{s p}=S \times(4S)^4  = {4 \times 4 \times 4 \times 4 S^5}$

- $K_{s p}=256S^5$

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-18.jpg)

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

- **Problem** The solubility of calcium sulphate in water is 4.9 $\times$ $10^{-3} mol {dm}^{-3}$ at 298 K. Calculate the value of Ksp for $CaSO_4$ at this temperature

- **Solution**

- $CaSO_4 \rightleftharpoons Ca^{2+}+SO_4^{2-}$

- $ \quad \quad \quad \quad   \quad \quad S \quad \quad S$

- $S = 4.9 \times 10^{-3}$

- $K_{s p} =S \times S=S^2=\left(4.9 \times 10^{-3}\right)^2$

</div>
	
<div style='float: right; width:0%;'>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-19.jpg)

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

- **Problem** The solubility product of a salt having general formula $MX_2$, in water is : $4 \times 10^{-12}$.The concentration of $M^{2+}$ ions in the aqueous solution of the salt is

- **Solution**

- $MX_2 \rightleftharpoons M^{2 +}+2 X^{-}$

- $ \quad \quad  \quad \quad \quad S \quad \quad S$

- $K_{s p}=S \times(2S)^2=S \times 4 S^2=4S^3$

- $4 \times 10^{-12}=4 \times S^3$

- $S=1 \times 10^{-4}$

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-20.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Problem with solution $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Factors affecting solubility $\rightarrow$ Problems with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problems with solution </primary>
<hr>
<main>

- **Problem** The $K_{sp}$ of $Ag_2 CrO_4$ is $1.1 \times 10^{-12}$ at 298 K. The solubility (in mol/l) of  $Ag_2CrO_4$ in a 0.1 M $AgNO_3$ solution is

- **Solution**

- $Ag_2 CrO_4 (s) \rightleftharpoons 2 Ag^{+}+CrO_4^{2-}$

- $\quad\quad\quad\quad\quad\quad\\quad\quad2S \quad\quad \quad S$

- $K_{SP}=[Ag+]^2[CrO_4^{2-}] =(2S)^2 \times S$

- $K_{SP} =4S^3$

- $4S^3 = 1.1 \times 10^{-12}$

</div>
	
<div style='float: right; width:0%;'>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-22.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Problem with solution $\rightarrow$ Factors affecting solubility $\rightarrow$ <span style = "color:blue"> Problems with solution </span> $\rightarrow$ Problems with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problems with solution </primary>
<hr>
<main>

- **Solution**

- $Ag_2 CrO_4 \rightleftharpoons 2 Ag^{+}+CrO_4^{2-}$

- $AgNO_3 \longrightarrow Ag^++NO_3^{-}$

- $\quad\quad\quad\quad\quad\quad  0.1 M \quad 0.1 M $

- $Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}$

- s= y mol/L  $\quad\quad$ 2y+0.1 $\quad\quad$ y

- $\quad\quad\quad\quad\quad\quad \approx$ 0.1

</div>
	
<div style='float: right; width:0%;'>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-23.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Factors affecting solubility $\rightarrow$ Problems with solution $\rightarrow$ <span style = "color:blue"> Problems with solution </span> $\rightarrow$ Problem with solution $\rightarrow$ Ionic product and solubility product </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

- **Solution**

- $K_{SP_{Ag_2CrO_4}}$=$[Ag^{+}]^2[CrO_4^{2-}]$

- $1.1 \times 10^{-12}=(0.1)^2 \times y$

- $y=\frac{1.1\times 10^{-12}}{[0.1]^2}$

- $y=1.1 \times 10^{-10} mol / L$

- $4S^3  =1.1 \times 10^{-12}$

- $S^3  =.25 \times 10^{-12}$

- $S \approx 15 \times 10^{-4}$

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-24.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Problems with solution $\rightarrow$ Problems with solution $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Ionic product and solubility product $\rightarrow$ Ionic product and solubility product </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Ionic product and solubility product </primary>
<hr>
<main>

-  Ionic product and solubility product for sparingly  soluble salt like AgCl

- $AgCl \rightleftharpoons [Ag^{+}][Cl^{-}]$
 
 - $[Ag^{+}][Cl^{-}]$ = $K_{salt}$ (Ionic product)

</div>

<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/04641877-691c-4a02-c4f1-a71415b38700/public" width="80%">

</main>
<hr>
<footer style = "font-size: 18px">Problems with solution $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Ionic product and solubility product </span> $\rightarrow$ Ionic product and solubility product $\rightarrow$ Application of the concept of solubility product </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Ionic product and solubility product </primary>
<hr>
<main>

-  AgCl $\rightleftharpoons Ag^+ + Cl^-$

- $\quad\quad\quad\quad$ **In solution**

- Multiplication of  $[Ag^{+}][Cl^{-}]$  in a saturated  solution is a known as solubility product

</div>

</main>
<hr>
<footer style = "font-size: 18px">Problem with solution $\rightarrow$ Ionic product and solubility product $\rightarrow$ <span style = "color:blue"> Ionic product and solubility product </span> $\rightarrow$ Application of the concept of solubility product $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Application of the concept of solubility product </primary>
<hr>
<main>

* Concept of solubility product can be used to know under what condition a precipitate will form

* Ionic product < solubility product then solution is not saturated

* Ionic product = solubility product then  solution is saturated

* Further addition of salt will lead to precipitation

*  Ionic product > solubility product then the  solution is oversaturated
 
</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-27.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Ionic product and solubility product $\rightarrow$ Ionic product and solubility product $\rightarrow$ <span style = "color:blue"> Application of the concept of solubility product </span> $\rightarrow$ Problem with solution $\rightarrow$ Problems with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

- **Problem** Solid Ba(NO<sub>3</sub>)<sub>2</sub> is gradually dissolved in a $ 1\times 10^{-4}M Na_2NO_3 $ solution.At what concentration of  $Ba^{2+} $ will a precipitate begin to form
- $(K_{sp}$ for $BaCO_3=5.1\times 10^{-9})$

- **Solution**

- $Na_2 CO_3 \longrightarrow 2 Na^{+}+CO_3^{2-}$

- $BaCO_3$  (sparingly soluble)

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-30.jpg)

</main>
<hr>
<footer style = "font-size: 18px">Ionic product and solubility product $\rightarrow$ Application of the concept of solubility product $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Problems with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

- **Problem** At $25^{\circ} {C}$, the solubility product of $Mg(OH)_2$ is $1.1\times 10^{-11}$. At which pH, will
- $Mg^{2+}$ ions start precipitating in the form of $Mg(OH)_2$ from a solution of $0.001 M Mg^{2+}$
- ions ?

- **Solution**

- $ K_{s p}=K_w$

- $ K_{SP}=[Mg^{2+}][OH^{-}]^2$

- $1.1 \times 10^{-11}=[Mg^{2+}][OH^{-}]^2$

- $[OH^-]^2=\frac{1.1 \times 10^{-11}}{.001}$

- $[OH^-]^2 = 1.1 \times 10^{-8}$

- $[OH^-] \approx 1 \times 10^{-4}$

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-33.jpg)

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

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

<div style='float: left; width:100%;font-size:30px;text-align:left'>
	
- **Problem** Calculate the molar solubility of $Mg(OH)_2$ in $1 MNH_4Cl$

- $K_{sp}Mg(OH)_2=1.8 \times 10^{-11}$

- $K_b(NH_3)=1.8 \times 10^{-5}$

- **Solution**

- $NH_4 Cl \rightarrow NH_4^{+}+Cl^{-}$

- $\quad 1M\quad\quad \quad\quad M $

- $NH_4^{+}+OH^{-} \rightleftharpoons NH_3+H_2O$

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-35.jpg)

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

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problems with solution </primary>
<hr>
<main>

- **Solution**

-  Suppose S is the solubility of $Mg(OH)_2$ in 1M $NH_4Cl$

- $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^{-}$

- $\quad\\quad\quad\quad\quad\quad\quad S \quad\quad\quad\quad 2S-X$

- $NH_4^{+} + OH^- \rightleftharpoons NH_3 + H_2O$

- $1 M \quad\quad2S  \quad\quad \quad0\quad\quad\quad \quad 0$

- $1-x   \quad 2S-x \quad\quad x \quad\quad\quad x$

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-37.jpg)

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

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
<hr>
<main>

- **Solution**

- $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^{-}$

- $ \quad \quad \quad \quad\quad \quad \quad S\quad\quad  \quad  2S-x$

- $NH_4^{+} + OH^- \rightleftharpoons NH_3 + H_2O$

- $\quad1-x \quad 2S-x \quad \quad \quad x$

- $\quad \approx1$

- $K_{SP} = S\times [2S-x]^2$

- $K_b=\frac{[NH_4^+[OH^-]}{[NH_3]}$
 - $=\frac{1 \times[OH^-]}{x}$

-  The concentration of $[OH^-]$ and S (solubility of salt) can be solved now

</div>

![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-08-ionic-equillibrium-l-8_8-n3q5ut39bva-38.jpg)

### <Success style="font-size:25px"> Ionic Equilibrium L-8 </Success>
### <primary style="font-size:24px"> Precipitation </primary>
<hr>
<main>

* If $Ag^+$ is added to a solution and we see a precipitate

* It is infer that $Cl^- \quad Br^- \quad I^-$ is present

</div>

<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/7e7f1189-868d-418d-8c5c-f375f23e4200/public" width="80%">

</main>
<hr>
<footer style = "font-size: 18px">Precipitation $\rightarrow$ Precipitation $\rightarrow$ <span style = "color:blue"> Precipitation </span> $\rightarrow$ Thank you $\rightarrow$  </footer>