Question

$Pb(I{O}_3{)}_2$ is a springly soluble salt $(K_{sp}=2.6\times {10}^{-13})$. To 35 mL of 0.15 M $Pb(N{O}_3{)}_2$ solution, 15 mL of 0.8 M $KI{O}_3$ solution is added, and a precipitate of $Pb(I{O}_3{)}_2$ is formed.

What will be the molarity of $I{O}_3^{⊝}$ ions in the solution after completion of the reaction?

• A. 0.152
• B. 0.081
• C. 0.41
• D. 0.03

Answer 1

The correct option is D 0.03

Number of millimoles of Pb(II) ions from lead nitrate $=35\times 0.15=5.25$. The number of millimoles of iodate ions from potassium iodate $=15\times 0.8=12$
1 mole of Pb ions will react with 2 moles of iodate ions. Hence, 5.25 millimoles of lead ions will react with 10.5 moles of iodate ions. The number of millimoles of iodate ions remaining unreacted $=12-10.5=1.5$. The molarity of the iodate ions $=\frac{1.5}{50}=0.03M$