Question

The concentration of Pb²⁺ in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of sodium carbonate, forming the insoluble lead (II) carbonate (267 g/mol) according to the balanced equation given below. The solid lead (II) carbonate is dried, and its mass is measured to be 0.1443 g. What was the concentration of Pb²⁺ in the original wastewater sample? Pb²⁺(aq) + Na₂CO₃(aq) ->PbCO₃(s) + 2Na⁺(aq)

Answer 1

Answer:

0.005404 M

Explanation:

$Pb^{2+}(aq) + Na_{2}CO_{3}(aq) ---> PbCO_{3}(s) + 2Na^{+}(aq)$

Since you added an excess of sodium carbonate you warrantied that all the $Pb^{+2}$ in the sample reacted with it. So we can say that the insoluble lead (II) carbonate $PbCO_{3}$ contains all the $Pb^{+2}$ ions in the original sample.  

The moles of $PbCO_{3}$ are:

$moles-of-PbCO_{3}=\frac{mass-of-PbCO_{3}}{Molecular-weight-of- PbCO_{3}}=\frac{0.1443g}{267\frac{g}{mol}}=0.00054 mol$

One mol of $Pb^{+2}$ is required to form one mol of $PbCO_{3}$. So, the stoichiometric relationship between them is 1:1.

$Pb^{+2} +CO_{3}^{-2} - - - > PbCO_{3}$

Knowing this, 0.00054 is also the number of moles of $Pb^{+2}$ in the original sample.  

So, the concentration of $Pb^{+2}$ in the original sample is:

$M = \frac{mol-of-Pb^{+2}}{volume-wastewater-(liters)}=\frac{0.00054}{0.1L}=0.005404 M$