Question

If 100 ml of 0,016 M CaCO3 solution is added to 50 ml of 0,01 M Pb(NO3)2

solution, will a precipitate form? What will it be the final concentration of dissolved Pb^2+ ?

Answer 1

Volume of CaCO3 solution, V1 = 100 mL = 100 mLx(1L/1000mL) = 0.1L

Concentration of CaCO3 solution, M1 = 0.016 M

Hence moles of CaCO3 in the solution = M1V1 = 0.016Mx0.1L = 1.6x10^-3  mol

Volume of Pb(NO3)2 solution, V2 = 50 mL = 50 mLx(1L/1000mL) = 0.05L

Concentration of Pb(NO3)2 solution, M2 = 0.01 M

Hence moles of Pb(NO3)2 in the solution = M2V2 = 0.01Mx0.05L = 5x10^-4 mol

Total volume of solution after mixng, Vt = 0.1L+0.05L = 0.15L

The concentration of Ca²⁺ (aq) and CO3^2-(aq) after mixing = moles of CaCO3 / Total volume

= 1.6x10^-3  mol / 0.15L = 1.1x10^-2 M

The concentration of Pb²⁺ (aq)and NO3- (aq) after mixing = moles of Pb(NO3)2 / Total volume

= 5x10^-4 mol / 0.15L = 3.33x10^-3 M

Since PbCO3 is less soluble than CaCO3, the following reaction occur when CaCO3 and Pb(NO3)2 are mixed.

Ca²⁺ (aq) + CO₃^2- (aq) + Pb²⁺ (aq)+ 2NO3-(aq) ------------------> Ca²⁺ (aq) + 2NO3- (aq) + PbCO3 (s)

Solubility product, Ksp of PbCO3 = 7.4x10^-14

Ionic product of PbCO3 = [Pb²⁺ (aq)]x[CO₃^2- (aq)] = 3.33x10^-3 Mx1.1x10^-2 M = 3.7x10^-5

Since ionic product > Ksp, precipitation of PbCO3 will occur. (answer)

In the balanced reaction 1 mol of CaCO3 reacts with 1 mol of Pb(NO3)2.

Since Pb(NO3)2 is present in smaller amount, it will be the limiting reagent.

Hence 5x10^-4 mol of Pb(NO3)2 will react with  5x10^-4 mol of CaCO3 to form  5x10^-4 mol of PbCO3.

Hence moles of CaCO3 left = 1.6x10^-3  mol - 5x10^-4 mol = 1.1x10^-3 mol

Hence [CO₃^2-(aq)] = 1.1x10^-3 mol / total volume =1.1x10^-3 mol/0.15L = 7.33x10^-3 M

Ksp =  7.4x10^-14 = [Pb²⁺ (aq)]x[CO₃^2- (aq)] = [Pb²⁺ (aq)]x7.33x10^-3 M

=> [Pb²⁺ (aq)] =  7.4x10^-14 /7.33x10^-3 M = 1.01x10^-11 M (answer)