You have two solutions, one \(1.50 \mathrm{M}\) sodium sulfide and the other \(1.00 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\). (a) Write a net ionic equation for the precipitation reaction that occurs when these solutions are combined. (b) How many milliliters of the solutions must be combined to prepare \(10.00 \mathrm{~g}\) of precipitate? (c) Suppose you filter off the precipitate and find that your percent yield is \(50.0 \%\). What volumes of the solutions should you have combined to isolate \(10.00 \mathrm{~g}\) of precipitate?

To find the net ionic equation, first, write the balanced chemical equation for the reaction between sodium sulfide and lead(II) nitrate. Then, separate the soluble ions in the equation, and finally, remove the spectator ions present on both sides of the equation. Balanced chemical equation: \(Na_2S(aq) + Pb(NO_3)_2(aq) \rightarrow PbS(s) + 2NaNO_3(aq)\) Now, separate the soluble ions: \(2Na^+(aq) + S^{2-}(aq) + Pb^{2+}(aq) + 2NO_3^-(aq) \rightarrow PbS(s) + 2Na^+(aq) + 2NO_3^-(aq)\) Next, remove the spectator ions (here, \(Na^+\) and \(NO_3^-\) ions): \(S^{2-}(aq) + Pb^{2+}(aq) \rightarrow PbS(s)\) This is our net ionic equation.

To find how many milliliters of solutions are needed to prepare 10 grams of precipitate, we need to use stoichiometry and the molar masses of the substances. First, we must find the molar mass of the precipitate, \(PbS\). Molar mass of \(PbS\) = \(207.2 g/mol (Pb) + 32.1 g/mol (S) = 239.3 g/mol\) Now, we will use stoichiometry to calculate the moles of \(PbS\) required: \(\frac{10.00 g PbS}{1} * \frac{1 mol PbS}{239.3 g PbS} \approx 0.0418 mol\) Since the stoichiometric ratio between \(Pb^{2+}\) and \(PbS\) is 1:1, we also need 0.0418 mol of \(Pb(NO_3)_2\) for this reaction. Let x be the volume (in mL) of the \(1.00 M Pb(NO_3)_2\) solution: \(1.00 M \times \frac{x mL}{1000} = 0.0418 mol\) Thus, \(x \approx 41.8 mL\) of the \(1.00 M Pb(NO_3)_2\) solution is needed. Now, we will do the same procedure for the \(1.50 M Na_2S\) solution. As the stoichiometric ratio between \(S^{2-}\) and \(PbS\) is 1:1, we need 0.0418 mol of \(Na_2S\) as well. Let y be the volume (in mL) of the \(1.50 M Na_2S\) solution: \(1.50 M \times \frac{y mL}{1000} = 0.0418 mol\) So, \(y \approx 27.87 mL\) of the \(1.50 M Na_2S\) solution is needed.

Since the percent yield is only 50%, we need to adjust the volumes of the solutions to obtain the desired amount of precipitate. If the yield is 50%, then we must double the amount of the reagents to obtain 10 grams of precipitate. So, we will double the volumes calculated in part (b): Volume of \(1.00 M Pb(NO_3)_2\) solution = \(41.8 mL \times 2 \approx 83.6 mL\) Volume of \(1.50 M Na_2S\) solution = \(27.87 mL \times 2 \approx 55.74 mL\) Hence, you should combine approximately 83.6 mL of the \(1.00 M Pb(NO_3)_2\) solution and 55.74 mL of the \(1.50 M Na_2S\) solution to isolate 10.00 g of precipitate with a 50% yield.