Answer:
0.005404 M
Explanation:
[tex]Pb^{2+}(aq) + Na_{2}CO_{3}(aq) ---> PbCO_{3}(s) + 2Na^{+}(aq)[/tex]
Since you added an excess of sodium carbonate you warrantied that all the [tex]Pb^{+2}[/tex] in the sample reacted with it. So we can say that the insoluble lead (II) carbonate [tex]PbCO_{3}[/tex] contains all the [tex]Pb^{+2}[/tex] ions in the original sample.
The moles of [tex]PbCO_{3}[/tex] are:
[tex]moles-of-PbCO_{3}=\frac{mass-of-PbCO_{3}}{Molecular-weight-of- PbCO_{3}}=\frac{0.1443g}{267\frac{g}{mol}}=0.00054 mol[/tex]
One mol of [tex]Pb^{+2}[/tex] is required to form one mol of [tex]PbCO_{3}[/tex]. So, the stoichiometric relationship between them is 1:1.
[tex]Pb^{+2} +CO_{3}^{-2} - - - > PbCO_{3}[/tex]
Knowing this, 0.00054 is also the number of moles of [tex]Pb^{+2}[/tex] in the original sample.
So, the concentration of [tex]Pb^{+2}[/tex] in the original sample is:
[tex]M = \frac{mol-of-Pb^{+2}}{volume-wastewater-(liters)}=\frac{0.00054}{0.1L}=0.005404 M[/tex]
