Question

If 12.8 g lead(II) sulfate (303.3 g/mol) precipitates when excess potassium chloride is added to 1.65 L of a water sample, what is molar concentration of Pb2+ in the sample?

Answer 1

Answer:

$M=0.0256M$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$PbSO_4(aq)+2KCl(aq)\rightarrow PbCl_2(s)+K_2SO_4(aq)$

In such a way, since all the lead (II) is converted due to the excess of potassium chloride, the moles of lead (II) in the sample are computed from the mass of lead (II) sulfate:

$n_{Pb^{2+}}=12.8gPbSO_4*\frac{1molPbSO_4}{303.3gPbSO_4} *\frac{1molPb^{2+}}{1molPbSO_4} \\\\n_{Pb^{2+}}=0.0422molPb^{2+}$

Thus, since volume of the solution is 1.65 due to the fact that the addition of the reactants is not enough to significantly modify the reaction's volume, the resulting molar concentration of the lead (II) ions is:

$M=\frac{n_{Pb^{2+}}}{V}=\frac{0.0422molPb^{2+}}{1.65L}\\ \\M=0.0256M$

Regards.