# The solubility product of  is . Calculate its solubility in gL-1 and also find out the pH of this solution. (Atomic mass of ).

As per the information given in the question,

Hence, we can write the equation of disassociation of 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,

Now, molar mass of

Therefore, solubility = molar mass x s

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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