Question

81.5 mL of a 3.30 M sodium chloride solution were used to react completely with 89.0 mL of an aqueous lead (II) nitrate solution. What is the molarity of the lead (II) nitrate solution?

Answer 1

Reaction

2NaCl+ Pb(NO₃)₂​ --------------->PbCl₂ + 2NaNO₃

molarity of Sodium chloride = (number of moles of sodium chloride)(1000)÷(volume of solution)

3.30M = (number of moles of sodium chloride)(1000)÷81.5

Number of moles of sodium chloride = 3.30M×81.5÷1000

Number of moles of sodium chloride = 0.26895mol

From the stoichiometry of the reaction

2 mole sodium chloride completely react with 1 mole of lead(ll) nitrate

So 1mole of sodium chloride completely react with 1/2 mole of lead (ll) nitrate

Hence 0.26895 mole of sodium chloride completely react with (1×0.26895)/2 mole of lead (ll) nitrate

Number of moles of lead (ll) nitrate = 1×0.26895/2

Number of moles of lead (ll) nitrate = 0.134475 mole

Molarity of Pb(NO₃) = {number of moles of Pb(No₃)}×1000÷volume of solution

Molarity of Pb(NO₃) = (0.134475mol)×1000÷89.0

Molarity of Pb(NO₃) = 1.5109M