Question

Solid sodium chloride is slowly added to 75.0 mL of a lead(II) nitrite solution until the concentration of chloride ion is 0.0232 M. The maximum amount of lead ion remaining in solution is _____M

Answer 1

Q98117

Solution:

As solid Sodium chloride $(NaCl)$ is added to the lead(II) nitrite solution, volume of the solution does not change effectively.

Thus, final moles of chloride ion $(Cl^-)=0.0232*75/1000$

$=1.74 milimoles.$ (Given)

According to the chemical reaction given below:

$2NaCl(aq)+Pb(NO_2)_2(aq) \to PbCl_2$ $(s)+2NaNO_2(aq)$

2 moles of Sodium chloride reacts with 1 mole of $Pb(NO_2)_2$ and 1 mole of it contains 1 mole of Pb2+ ions.

Thus, 2 moles of $Cl^-$ reacts with 1 mole of $Pb^{2+}$ ions.

Hence, final concentration of lead ions is

$=1.74/2 milimoles.$

$=0.87 milimoles.$

$=0.87*10^{-3}/(75/1000)M$

$=0.0116 M.$ (Answer)