Question

Upon mixing 45.0 mL 0.25 M lead nitrate solution with 25.0 mL of 0.10 NI chromic sulphate solution. Precipitation of lead sulphate takes place. How many moles of lead sulphate are formed ? Also calculate the molar concentrations of the species left behind in the final solution. Assume that lead sulphate is completely insoluble.

Answer 1

Pb(NO₃)₂ reacts with chromic sulphate as follows :

According to above equation :

1 mole of Pb²⁺ reacts with one mole of SO^2-₄ So conc. of Pb²⁺ is in excess and concentration of SO^2-₄ is in limit.

Millimoles of PbSO₄ : Millimoles of SO^2-₄

Remaining millimoles of Pb²⁺ = 17.25 - 7.5 = 3.75

No. of millimoles of NO^-₃ in the solution = 22.5

No. of millimoles of Cr³⁺ in the solution = 5.0

Total volume of the solution = 45.0 + 25.0 = 70.0 mL