Upon mixing 45.0 mL of 0.25 M lead nitrate solution with 25 mL of 0.1 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.

Upon mixing 45.0 mL of 0.25 M lead nitrate solution with 25 mL of 0.1

1.	Upon mixing 45.0 mL of 0.25 M lead nitrate solution with 25 mL of 0.1 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» Solution :The reaction is : `underset(3"mole")(3Pb(NO_(3))_(2))+underset(1"mole")(Cr_(2)(SO_(4))_(3)) to underset(3"mole")(3PbSO_(4))+underset(2"mole")(2Cr(NO_(3))_(3))` No. of moles of `Pb(NO_(3))_(2)=45xx10^(-3)xx0.25` `=11.25xx10^(-3)` mole No. of moles of `Cr_(2)(SO_(4))_(3)=25xx10^(-3)xx0.1=2.5xx10^(-3)` mole Thus, `Cr_(2)(SO_(4))_(3)` has limiting concentration. It shall be consumed fully and the number of moles of lead sulphate produced will be `=3xx2.5xx10^(-3)=7.5xx10^(-3)` mole No. of moles of lead nitrate left `=11.25xx10^(-3)-7.5xx10^(-3)` `=3.75xx10^(-3)` mole Total volume =(45.0+25.0)=70 ML of `70xx10^(-3)` litre Molarity `=(3.75xx10^(-3))/(70xx10^(-3))=0.0536 M` No. of moles of `Cr(NO_(3))_(3)` formed`=2xx2.5xx10^(-3)=5xx10^(-3)` mole Molarity `=(5xx10^(-3))/(70xx10^(-3))=0.0714 M` `Pb(NO_(3))_(2) " and " Cr(NO_(3))_(3)` will be PRESENT in solution in IONIC form. Thus, `[Pb^(2+)]=0.0536 M` `[Cr^(3+)]=0.0714 M` `[NO_(3)^(-)]=(2xx0.0536)+(3xx0.0714)` =0.3214 M