1.

The solubility product of BaSO4 is 1.5 x 10-9 . Find out its solubility in pure water.

Answer»

BaSO4 = Ba2+ + SO42-

Ks = (S)2 = 1.5 x 10-9

 S = \(\sqrt{1.5\times10^{-9}}\)  = \(\sqrt{1.5\times10^{-10}}\)

= 3.87 x 10-15 mol L-1



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