Question

5.0mg of NaF is added to 100ml of a 1.40*10^-2M solution of Pb(NO3)2.Calculate the resulting concentration of Pb2+ after mixing ?

Answer 1

When NaF is added to Pb(NO₃)₂ solution,soluble NaNO3 and insoluble PbF2 are formed. Hence PbF2 is precipitated as soon as it is formed. The balanced chemical reaction is

2NaF + Pb(NO₃)₂  -----------------> 2NaNO3 + PbF₂

Moles of NaF = 5.0x10^-3g/ 42 = 0.119x10^-3 mol = 0.119 mmol.

moles of Pb(NO₃)₂ = MxV = 1.4x10^-2x0.100L = 1.4x10^-3 mol = 1.4 mmol

Since 2 moles of NaF reacts with 1 mol of  Pb(NO₃)₂ , here NaF is the limiting reagent.

0.119 mmol of NaF will react with 0.119mmol/2 = 0.0595 mmol of Pb(NO₃)_2.

Hence moles of Pb(NO₃)₂ left in the solution = (1.4 - 0.0595)mmol = 1.3405 mmol

Since Pb(NO₃)₂ is completely ionised, hence moles of Pb²⁺ in the solution = 1.3405 mmol

Total volume of the solution = 100 mL = 0.100L

Therefore concentration of Pb²⁺ after mixing = 1.3405 x10^-3 mol/ 0.100L = 0.013405M = 1.34 x10^-2 M (answer)