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Question

Chemistry Question on Equilibrium

What is the molar solubility of Al(OH)3Al(OH)_3 in 0.2 M NaOH0.2\, M\, NaOH solution ? Given that, solubility product of Al(OH)3=2.4×10−24Al(OH)_3 = 2.4 \times 10^{-24} :

A

12×10−23 12 \times 10^{- 23}

B

12×10−21 12 \times 10^{- 21 }

C

3×10−19 3 \times 10^{- 19}

D

3×10−22 3 \times 10^{- 22}

Answer

3×10−22 3 \times 10^{- 22}

Explanation

Solution

Al(OH)3<=>Al+3+3OH−{ Al(OH)_3 <=> Al^{+3} + 3OH^{-}}
S′  0.2+3(S′)≃0.2S' \; 0.2 + 3 (S') \simeq 0.2
S;×(0.2)3=ksp=2.4×10−24S; \times (0.2)^3 = k_{sp} = 2.4 \times 10^{-24}
(S′)=3×10−22M(S') = 3 \times 10^{-22} M