0.2 mole of MgCl 2 (s) are added in 1 litre volume of a solution (S), already containing 0.2 mole of NaOH(s). Now answer the following questions :
(i) Calculate pH of obtained solution. K sp of Mg(OH) 2 is 1.6 × 10 –12 .
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
As the value of K sp is very low so we can assume that almost all the Mg(OH) 2 will Present in solid state.
(i) Mg +2 + 2OH – 
Mg(OH) 2
t = 0 0.2 0.2 0
0.1 x 0.1
At the end of reaction [Mg +2 ] = 0.1
K sp [Mg (OH) 2 ] = [Mg +2 ] [OH – ] 2 = 1.6 × 10 –12
⇒ [OH – ] 2 = 
⇒ [OH – ] = 4 × 10 –6 M
⇒ pOH = 6 – log 4 ⇒ pH = 14 – 6 + 0.6 = 8.6
(ii) Mg +2 + 2OH – 
Mg (OH) 2 (s)
0.1 0.04
At the end of reaction [Mg +2 ] = 0.1 – 
= 0.08
K sp (Mg(OH) 2 ) = [Mg +2 ] [OH – ] 2 = 1.6 × 10 –12
[OH – ] 2 = 
⇒ [OH – ] = 4.47 × 10 –6
∴ pOH = 6 – log 4.47 = 5.35
pH = 14 – 5.35 = 8.65
(iii) Mg(OH) 2 + 2HCI 
Mg +2 + 2CI – + 2 H 2 O
0.1 0.04 0.1 - -
0.1 – 0.02 0 0.1 + 0.02
[Mg +2 ] = 0.12 M
K sp [Mg (OH) 2 ] = [Mg +2 ] [OH – ] 2 = 1.6 × 10 –12
[OH – ] 2 = 
= 
× 10 –11 M 2 ⇒ [OH – ] = 
M
⇒ pOH = 5.44 ⇒ pH = 14 – 5.44 = 8.56
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