The solubility product of Al(OH)3 is 2.7 × 10–11. Calculate its solubility in g L–1 and also find out pH of this solution. (Atomic mass of Al is 27 u).

Let S be the solubility.


Initially, assume 1 mole of Al(OH)3 was present. From which S moles of it was dissolved to give S and 3S moles of Al3+ and 3OH- respectively. This can be understood from the following equation.



Given, = 2.7 × 10-11


We know,


= [Al3+][OH-]3


= (S)(3S)3 = 27S4


S4 = = 1 ×


S = 1 × 10-3 mol L-1


Molar mass of Al(OH)3 = 27 + 163 + 13 = 78 g/mol


Solubility of Al(OH)3 in g L-1 = 1 × × 78 g/L


= 78 × = 7.8 × 10-2 g L-1


Hence, solubility of Al(OH)3 is 7.8 × 10-2 g L-1.


Now to find pH of the solution:


From above equation,


[OH-] = 3S = 3 × 1 × = 3 ×


We know,




We also know, pH + pOH = 14



Therefore, pH of the solution is 11.47


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