Question 13.7: Consider the precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq)......
Consider the precipitation reaction:
2\operatorname{KI}(a q)+\operatorname{Pb}(\operatorname{NO}_{3})_{\mathbb{2}}(a q)\longrightarrow\operatorname{PbI}_{2}(s)+2\operatorname{KNO}_{3}(a q)How much 0.115 M KI solution in liters will completely precipitate the Pb^{²+} in 0.104 L of 0.225 M Pb(NO_3)_2 solution?
GIVEN: 0.115 M KI
0.104 L Pb(NO_3)_2 solution
0.225 M Pb(NO_3)_2
FIND: L KI solution
RELATIONSHIPS USED
M KI =\frac{ 0.115 mol KI}{L KI solution } (given molarity of Pb(NO_3)_2 solution, written out as a fraction)
M Pb(NO_3)_2= \frac{0.225\,\mathrm{mol\;Pb}(\mathrm{NO}_{3})_{2}}{\mathrm{L\;Pb}(\mathrm{NO}_{3})_{2}\,\,{\mathrm{solution}}} (given molarity of Pb(NO_3)_2 solution, written out as a fraction)
2 mol KI K ≡ 1 mol Pb(NO_3)_2 (stoichiometric relationship between KI and Pb(NO_3)_2, from balanced chemical equation)
0.104 \cancel{L Pb(NO_3)_2 \mathrm{solution}} \times \frac{0.225 \cancel{mol Pb(NO_3)_2}}{\cancel{L Pb(NO_3)_2 \mathrm{solution}}} \times \frac{2 \cancel{mol KI}}{ \cancel{mol Pb(NO_3) }} \\ \times \frac{ L KI \mathrm{solution}}{0.115 \cancel{mol KI}}=0.407 L KI \mathrm{solution}The units (L KI solution) are correct. The magnitude of the answer makes sense because the lead nitrate solution is about twice as concentrated as the potassium iodide solution and 2 mol of potassium iodide are required to react with 1 mol of lead(II) nitrate. Therefore we would expect the volume of the potassium solution required to completely react with a given volume of the Pb(NO_3)_2 solution to be about four times as much.
