Solubility product

source : toppr

## Solubility product Numerical-

### Question 1) The solubility of Mg(OH)_{2} in a particular buffer is found to be 0.95 gm/ litre .What is the pH of the buffer solution? *K *_{sp} of Mg(OH)_{2} = 1.8x 10^{-11}

_{sp}of Mg(OH)

_{2}= 1.8x 10

^{-11}

### Solution )

*S of Mg(OH) _{2} = 0.95 gm /litre*

*mol.wt. of Mg(OH) _{2} = 24 + (16 + 1)2 = 24 + 34 = 58*

*S in mole /litre = ( S in gm /litre) / mol.wt.= 0.95 / 58*

*S = 0.01637 = 0.0164 mole/litre*

*Mg(OH) _{2} ⇌ Mg^{++} + 2 OH^{–}*

* K _{sp} = [Mg^{++}][OH^{–}]^{2}*

*1.8 x 10 ^{-11} = 0.0164 [OH^{–}]^{2}*

*[OH ^{–}]^{2} = 1.8 x 10^{-11} / 0.0164 = 109.75 x 10^{-11} =10.975 x 10^{-10}*

*[OH ^{–}] = √ 10.975 x 10^{-10} = 3.31 x 10^{-5} mole/litre*

*pOH = – log [OH ^{–}]= -log [3.31 x 10^{-5}]*

*= -log 3.31 + 5 log 10= -0.5198 + 5*

*pOH = 4.48*

*pH = 14 – pOH= 14 – 4.48*

*pH =9.52 Ans.*

### Question 2) Equal volumes of 0.02 M CaCl_{2} and 0.0004 M Na_{2}SO_{4} are mixed . Will a precipitate form ?

### ( K_{sp} of CaSO_{4} = 2.4 x 10^{-5} )

### Solution )

* CaCl _{2} + Na_{2}SO_{4} ——–> CaSO_{4} + 2NaCl*

*Milli moles 0.02V 0.0004V 0 0*

*Assume, V ml of both are mixed.*

*Total volume = V + V = 2V*

*[Ca ^{++}] = 0.02 V/2V = 0.01*

*[SO4 ^{—}] = 0.0004 V/2V = 0.0002 = 2 x 10^{-4}*

*ionic product of [Ca ^{++}] [SO_{4}^{—}] =0.01 x 2 x 10^{-4}= 2 x 10^{-6}*

*K _{sp} of CaSO_{4} = 2.4 x 10^{-5}*

*If ionic product of [Ca ^{++}] [SO_{4}^{—}] is greater than K _{sp} of CaSO_{4} , then precipitation takes place But,*

* ionic product of [Ca ^{++}] [SO_{4}^{—}] < K _{sp} of CaSO_{4}*

* 2 x 10 ^{-6} < 2.4 x 10^{-5}*

*Therefore CaSO*_{4} will not precipitate.

_{4}will not precipitate.

### Question 3) A sample of hard water contains 0.005 mole of CaCl_{2} per litre. What is the minimum conc. of Na_{2}SO_{4} which must be added for removing Ca^{++} ions from this water sample? K _{sp} of CaSO_{4} = 2.4 x 10^{-5} at 25^{0}C ?

### Solution )

*K _{sp} of CaSO_{4} = 2.4 x 10^{-5}*

*CaSO _{4} ⇌ Ca^{++} + SO_{4}^{—}*

*[SO _{4}^{—}] = y mole/litre is sufficient to precipitate CaSO_{4} from a solution having ,*

*[Ca ^{++}] = 0.005 mole /litre*

*K _{sp} = [Ca^{++}] [SO_{4}^{—}]*

*2.4 x 10 ^{-5} = 0.005 y*

*y = 2.4 x 10 ^{-5} / 0.005= 0.48 x 10^{-2}*

*y = 4.8 x 10 *^{-3} mole/litre

^{-3}mole/litre

*Minimum conc. of Na*_{2}SO_{4} which must be added for removing Ca^{++} ions from sample = 4.8 x 10 ^{-3} mole/litre

_{2}SO

_{4}which must be added for removing Ca

^{++}ions from sample = 4.8 x 10

^{-3}mole/litre

### Question 4) The *Solubility product *of PbBr_{2} is 8 x 10^{-5}. if the salt is 80 % dissociated in saturated solution. Find the solubility of salt in gm / litre ?

### Solution )

*Suppose, solubility of salt = S mole / litre*

*PbBr _{2} ⇌ Pb^{++} + 2 Br^{–}*

* 0.80 S 2 x 0.80 S*

*K _{sp} = [Pb^{++}] [Br^{–}]^{2}*

*Solubility product of PbBr _{2} = 8 x 10^{-5}*

*8 x 10 ^{-5} = 0.80 S ( 2 x 0.80 S)^{2}*

*S ^{3} = 3.906 x 10^{-5} = 39.06 x 10^{-6}*

*S = 3 √ 39.06 x 10^{-6}*

*S = 3.39 x 10*^{-2} mole /litre

^{-2}mole /litre

*mol.wt of PbBr _{2} = 207 + 2 x 80 = 367*

*S in gm /litre = (S in mole /litre) x mol.wt.= 3.39 x 10 ^{-2} x 367*

< h4>

*S = 12.44 gm / litre Ans.*