Question

Upon mixing \[45.0{\text{ }}ml\] of \[0.25{\text{ }}M\] lead nitrate solution with \[25{\text{ }}ml\]of \[0.1{\text{ }}M\] chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also, calculate the molar concentrations of the species left behind in the final solution. Assume that lead sulphate is completely insoluble.

Answer 1

Hint: One must know what a limiting reagent is in a reaction, a limiting reagent in a chemical reaction is a reactant that is totally consumed or used up when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without the limiting reactant.

Complete answer:
Now the general balanced reaction for the formation of lead sulphate is:
\[3Pb{\left( {N{O_3}} \right)_2} + C{r_2}{\left( {S{O_4}} \right)_3} \to 3PbS{O_4} + 2Cr{\left( {N{O_3}} \right)_3}\]
Now as we are given with the molarity and the volume of the reactants, the number of moles of reactants can be calculated easily as we need the number of moles to detect the limiting reagent present in the reaction.
$no.of moles = Molarity \times volume(litres)$
Number of moles of \[Pb{\left( {N{O_3}} \right)_2}\]can be calculated as:
\[ = 0.25 \times 45 \times {10^{ - 3}}\]
\[ = 11.25 \times {10^{ - 3}}mole\]
And Number of moles of \[C{r_2}{\left( {S{O_4}} \right)_3}\]can be calculated as:
\[ = 0.1 \times 25 \times {10^{ - 3}}\]
\[ = 2.5 \times {10^{ - 3}}mole\]
Thus, \[C{r_2}{\left( {S{O_4}} \right)_3}\]​ has a limiting concentration, as it will be consumed fully and the number of moles of lead sulphate produced will totally depend on the moles of \[C{r_2}{\left( {S{O_4}} \right)_3}\], now as in the general reaction for one mole of \[C{r_2}{\left( {S{O_4}} \right)_3}\] we obtain 3 moles of PbSO$_4$. Thus, multiplying the moles of \[C{r_2}{\left( {S{O_4}} \right)_3}\] with \[3\] we will get the moles of \[PbS{O_4}\] formed:
\[ = 3 \times 2.5 \times {10^{ - 3}}\]
\[ = 7.5 \times {10^{ - 3}}mole\]
Now, the lead nitrate left in the reaction can be easily calculated by subtracting the reacted amount:
Number of moles of lead nitrate \[Pb{\left( {N{O_3}} \right)_2}\] left,
\[ = 11.25 \times {10^{ - 3}} - 7.5 \times {10^{ - 3}} = 3.75 \times {10^{ - 3}}mole\] .
Now, as we have calculated the no. of moles of lead nitrate \[Pb{\left( {N{O_3}} \right)_2}\] left, we can calculate its molality if we calculate the total volume which is,
\[Total{\text{ }}volume\; = \left( {45.0 + 25.0} \right) = 70\;mL\;or\;70 \times {10^{ - 3}}litre\]
And molarity of Lead nitrate \[Pb{\left( {N{O_3}} \right)_2}\]left is,
\[Molarity = \dfrac{{3.75 \times {{10}^{ - 3}}}}{{70 \times {{10}^{ - 3}}}} = 0.0536\;M\]
And, Number of moles of $Cr{(N{O_3})_3}$ formed,
\[ = 2 \times 2.5 \times {10^{ - 3}} = 5 \times {10^{ - 3}}mole\]
It is multiplied by $2$ as $2$ moles of the $Cr{(N{O_3})_3}$ is formed in the general reaction, And Molarity of $Cr{(N{O_3})_3}$ is,
\[\; = \dfrac{{5 \times {{10}^{ - 3}}}}{{70 \times {{10}^{ - 3}}}} = 0.0714\;M\]
As in the question it is given that lead sulphate \[PbS{O_4}\] is incompletely soluble then it won’t exist as ions but, \[Pb{\left( {N{O_3}} \right)_2}\] and $Cr{(N{O_3})_3}$​ will be present in solution in ionic form.
Thus, concentration of $[P{b^{2 + }}]$ will be equal to \[0.0536\;M\], and the concentration of chromium ion \[[C{r^{3 + }}]\]is equal to \[0.0714\;M\], and as nitrate ion is coming from \[Pb{\left( {N{O_3}} \right)_2}\] and $Cr{(N{O_3})_3}$ as well hence concentration of $[N{O_3}^ - ]$ will be equal to:
\[\left( {2 \times 0.0536} \right) + \left( {3 \times 0.0714} \right) = 0.3214\;M.\]

Note:
Now a student can confuse that the reacted amount of limiting reagent \[C{r_2}{\left( {S{O_4}} \right)_3}\] is \[2.5 \times {10^{ - 3}}mole\] then why the reacted amount of \[Pb{\left( {N{O_3}} \right)_2}\] is \[7.5 \times {10^{ - 3}}mole\], the reason is because three moles of \[Pb{\left( {N{O_3}} \right)_2}\] are reacting with one moles of \[C{r_2}{\left( {S{O_4}} \right)_3}\] hence, the reacted amount of \[Pb{\left( {N{O_3}} \right)_2}\] is thrice of \[C{r_2}{\left( {S{O_4}} \right)_3}\].