The solubility product of $BaSO_4$ is $1.5 \times 10^{-9}$. The precipitation in a $0.01 \ M \ Ba^{2+}$ solution will start,on adding $H_2SO_4$ of concentration:
- A$10^{-9} \ M$
- B$10^{-8} \ M$
- C$10^{-7} \ M$
- D$10^{-6} \ M$
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