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Question
calculate the molar solubility of Ni(OH)2 in 0.1 M NaOH . The ionic product of Ni(OH)2 is 2 x 10-15
calculate the molar solubility of Ni(OH)2 in 0.1 M NaOH . The ionic product of Ni(OH)2 is 2 x 10-15
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Solution
Ni(OH)2 (s) <===> Ni2+ (aq) + 2 OH¯ (aq)
Ionic product expression: :
Ionic product =[Ni2+][OH-]2
let molar solubility of Ni2+ be x M
then concentration of OH- =(2x + 0.1) M
​2 x 10-15 = x.(2x + 0.1)2
Since x is very very small as compared to 0.1M thus ignoring 2x
​x<<<0.1
2x10-15 =0.01x
x =2 x 10-17 M
​Molar solubility of Ni(OH)2 is 2 x 10-17 M​
Ionic product expression: :
Ionic product =[Ni2+][OH-]2
let molar solubility of Ni2+ be x M
then concentration of OH- =(2x + 0.1) M
​2 x 10-15 = x.(2x + 0.1)2
Since x is very very small as compared to 0.1M thus ignoring 2x
​x<<<0.1
2x10-15 =0.01x
x =2 x 10-17 M
​Molar solubility of Ni(OH)2 is 2 x 10-17 M​
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