First slide

Solubility product(KSP)

Question

Calculate the molar solubility of $Ni (OH)_{2}$ in 0.10 M NaOH. The ionic product of $Ni (OH)_{2}$ is $2.0 \times 10^{- 15}$ .

Difficult

A

$6.0 \times 10^{- 12} M$
B

$8.0 \times 10^{- 13} M$
C

$2.0 \times 10^{- 13} M$
D

$5.0 \times 10^{- 12} M$

Solution

Let the solubility of $Ni (OH)_{2}$ = S, dissolution of S mol/L of $Ni (OH)_{2}$ provides S mol/L of ${Ni}^{2 +}$ and 2S mol/L of ${OH}^{-},$
but the total concentration of ${OH}^{-}$ = (0.10 + 2S ) mol/L because the solution already contains 0.10 mol/L of $\bar{O} H$ from NaOH.
$K_{sp} = 2.0 \times 10^{- 15} = [{Ni}^{2 +}] {[{OH}^{-}]}^{2}$
$= (S) (0.10 + 2 S)^{2}$
$As K_{sp} is small, 2 S < 0.10, thus, (0.10 + 2 S) \approx 0.10$
Hence, $2.0 \times 10^{- 15} = S (0.10)^{2}$
$S = 2.0 \times 10^{- 13} M = [{Ni}^{2 +}]$

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