Learning Objectives Determine amounts of reactants or products in aqueous solutions. As we learned in Chapter 7, double replacement reactions involve the reaction between ionic compounds in solution and, in the course of the reaction, the ions in the two reacting compounds are “switched” (they replace each other). Because these reactions occur in aqueous solution, we can use the concept of molarity to directly calculate the number of moles of reactants or products that will be formed, and hence their amounts (i.e. volume of solutions or mass of precipitates). As an example, lead(II) nitrate and sodium chloride react to form sodium nitrate and the insoluble compound, lead(II) chloride. \[\ce{Pb(NO_3)_2 (aq) + 2 NaCl (aq) → PbCl_2 (s) + 2 NaNO3 (aq)}\] In the reaction shown above, if we mixed .123 L of a 1.00 M solution of NaCl with 1.50 M solution of Pb(NO 3 ) 2 , we could calculate the volume of Pb(NO 3 ) 2 that reacted to completely precipitate the Pb 2 + ions. The molar concentration can also be expressed as the following: 1.00M NaCl = \(\frac{1.00 \; mol \; NaCl}{1 \; L \; NaCl \; solution}\) and 1.50M Pb(NO 3 ) 2 = \(\frac{1.00 \; mol \; Pb(NO_3)_2}{1 \; L \; Pb(NO_3)_2 \; solution}\) First, we must examine the reaction stoichiometry. In this reaction, one mole of Pb(NO 3 ) 2 reacts with two moles of NaCl to give one mole of PbCl 2 . Thus, the concept map utilizing the stoichiometric ratios is: so the volume of lead (II) nitrate that reacted is calculated as: \[.123 \; L \; NaCl solution \times \frac{1 \; L \; NaCl \; solution}{1.00 \; mol \; NaCl} \times \frac{1 \; mol \; Pb(NO_3)_2}{2 \; mol \; NaCl} \times \frac{1 \; L \; Pb(NO_3)_2 \;solution}{1 \; mol \; Pb(NO_3)_2} = 0.4 \; Pb(NO_3)_2 \; solution\] Example \(\PageIndex{1}\): What volume (in L) of 0.500 M sodium sulfate will react with 275 mL of 0.250 M barium chloride to completely precipitate the Ba 2 + in the solution? Solution Steps for Problem Solving Example \(\PageIndex{1}\) Identify the "given"information and what the problem is asking you to "find." Given: 275 mL BaCl2 0.250 M BaCl2 or \(\frac{0.250\; mol BaCl_2}{1\; L\; BaCl_2\; solution}\) 0.500 M Na2SO4 or \(\frac{0.500\; mol Na_2SO_4}{1\; L\; Na_2SO_4\; solution}\) Find: Volume Na2SO4 Set-up and balance the chemical equation Na2SO4(aq) + BaCl2(aq) BaSO4(s) + 2 NaCl (aq) An insoluble product is formed after the reaction. List other known quantities 1 mol of Na2SO4 to 1 mol BaCl2 1000 mL = 1 L Prepare a concept map and use the proper conversion factor. NaN Cancel units and calculate. \(275\cancel{mL \; BaCl_2 \; solution}\times \frac{1\cancel{L}}{1000\cancel{mL}}\times \frac{0.250 \cancel{mol \; BaCl_2}}{1 \cancel{BaCl_2 \; solution}}\times \frac{1 \cancel{mol \; Na_2SO_4}}{1 \cancel{mol \; BaCl_2}}\times \frac{1 L \; Na_2SO_4 \; solution}{0.500 \cancel{mol Na_2SO_4}}\) Think about your result. The lesser amount (almost half) of sodium sulfate is to be expected as it is more concentrated than barium chloride. Also, the unit is correct. Exercise \(\PageIndex{1}\) What volume of 0.250 M lithium hydroxide will completely react with 0.500 L of 0.250 M of sulfuric acid solution? Answer: 0.250 L LiOH solution