Question

What is the concentration of NO3− ions in the reaction solution formed by mixing 35.8 ml of aqueous 0.255 M Pb(NO3)2 with 19.1 ml of 0.415 M NaCl, given the balanced chemical equation-

Pb(NO₃)₂(aq)+2NaCl(aq)→PbCl₂(s)+2NaNO₃(aq)
A) 0.300M
B) 0.170M
C) 0.085M
D) 0.255M

Answer 1

The concentration of NO3− ions in the reaction solution is 0.300M.

Step-by-step explanation:

The molar concentration of NO3− ions in the reaction solution can be calculated using the stoichiometry of the balanced chemical equation and the given volumes and concentrations of the reactants.

First, calculate the moles of Pb(NO3)2 and NaCl using the given volumes and concentrations:

Moles of Pb(NO3)2 = volume of Pb(NO3)2 solution x concentration of Pb(NO3)2 = 35.8 ml x 0.255 M = 9.129 mol

Moles of NaCl = volume of NaCl solution x concentration of NaCl = 19.1 ml x 0.415 M = 7.9165 mol

Since the stoichiometry of the balanced chemical equation is 1:2 for Pb(NO3)2:NaCl, the moles of NO3− ions in the reaction solution will be twice the moles of NaCl:

Moles of NO3− ions = 2 x moles of NaCl = 2 x 7.9165 mol = 15.833 mol

Finally, calculate the concentration of NO3− ions in the reaction solution by dividing the moles by the total volume of the solution:

Concentration of NO3− ions = moles of NO3− ions / total volume of solution = 15.833 mol / (35.8 ml + 19.1 ml) = 0.300 M