Ncert Exercises & Solutions

The solubility product of $Al {(OH)}_{3}$ is $2.7 \times 10^{- 11}$ . Calculate its solubility in g L^-1 and also find out pH of this solution. (Atomic mass of Al=27 u)

Let S be the solubility of

Al {(OH)}_{3}

\underset{1 - s}{\underset{1}{Al {(OH)}_{3}}} ⇌ \underset{s}{\underset{0}{{Al}^{3 +}}} (aq) + \underset{3 s}{\underset{0}{3 {OH}^{-}}} (aq)

Concentration of species at t=0

Concentration of various species at equilibrium

K_{sp} = [{Al}^{3 +}] {[{OH}^{-}]}^{3} = (S) {(3 S)}^{3} = 27 S^{4}

S^{4} = \frac{K_{sp}}{27} = \frac{2.7 \times 10^{- 11}}{27} = 1 \times 10^{- 12}

S = 1 \times 10^{- 3} mol L^{- 1}

(i) Solubility of $Al {(OH)}_{3}$

Molar mass of

Al {(OH)}_{3}

is 78 g. Therefore,

Solubility of

Al {(OH)}_{3}

g L^{- 1} = 1 \times 10^{- 3} \times 78 g L^{- 1} = 78 \times 10^{- 3} g L^{- 1} = 7.8 \times 10^{- 2} g L^{- 1}

(ii) pH of the solution

S = 1 \times 10^{- 3} mol L^{- 1}

[{OH}^{-}] = 3 S = 3 \times 1 \times 10^{- 3} = 3 \times 10^{- 3}

pOH = 3 - \log 3

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