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

Answers (1)

As per the information given in the question,

K_{sp} =2.7\times 10^{-11}

Hence, we can write the equation of disassociation of Al(OH)_{3} as -

Al(OH)_{3}\rightleftharpoons Al^{3+}(aq)+3OH^{-}(aq)

At t=0 1 0 0
At,t=t 1-s s 3s

It is known to us that,

K_{sp}= [Al^{3+}] [OH^{-}] ^{3-}

=(s)\times (3s)^{3}

=27s^{4}

S^{4}=\frac{K_{sp}}{27}

S^{4}=\frac{2.7\times 10^{-11}}{27}

S^{4}=10^{-12}

S=(10^{-12})^{\frac{1}{4}}=10^{-3}mol/L

Now, molar mass of Al(OH)_{3}=78

Therefore, solubility = molar mass x s

=78\times10^{-3}

=7.8\times10^{-2}g/L

Now, it is known to us that,

pH = 14 - pOH

[OH]= 3s = 3 × 10-3

pOH= 3-log3

pH = 14 – 3 + log3

= 11.4771

Therefore, we can conclude that the solubility in g/L will be 7.8×10-2 g/L amd the pH of the solution will be 11.47.

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