Question

Many lead salts are often used as pigments. If $PbS{O}_4$ $(K_{sp}=1.6\times {10}^{-8})$ were used in an unglazed ceramic bowl, how many milligrams of lead (II) could dissolve per liter of water?

• A. 43
• B. 35
• C. 11
• D. 38

Answer 1

The correct option is D 38

The relationship between solubility and solubility product for lead sulphate is $S=\sqrt{K_{sp}}=\sqrt{1.6\times {10}^{-8}}=1.26\times {10}^{-4}M$

The molar mass of lead sulphate is $303.26g/mol$.

The solubility in g/L is $303.26\times 1.26\times {10}^{-4}=3.83\times {10}^{-2}g/L=38mg/L$

Thus 38 milligrams of lead (II) could dissolve per litre of water.

Hence, option D is correct.