Calculations Involving Concentrations

The stoichiometry calculations done in Chapter 8 involved pure substances and samples whose mass could be determined by weighing. In Chapter 9, the reactant was often a gas. In these cases the mass of the sample could be determined if we knew its pressure, volume, and temperature by applying the ideal gas equation and the concept of molar volume. Now, by using the concentration of a solution as a conversion factor, we can extend stoichiometric calculations to include solutions.

 Example: What weight of barium sulfate is precipitated by the addition of an excess of sulfuric acid to 55.6 mL of 0.54 M barium chloride? Solution Equation BaCl2 + H2SO4 BaSO4 + 2 HCl Given 55.6 mL of 0.54 M BaCl2 Conversion factors 0.54 mol BaCl2 in 1 L of 0.54 M BaCl2 0.54 mmol BaCl2 in 1 mL of 0.54 M BaCl2 1 mol BaCl2 forms 1 mol BaSO4 1 mol BaSO4 weighs 233.4 g Arithmetic equation Answer 7.0 g BaSO4

In the next example, two solutions are used. The concentration and volume of the first are known; only the concentration of the second is known. Its volume can be calculated.

 Example: What volume of 0.154 M sodium hydroxide will completely react with 25.0 mL of 0.0952 M hydrochloric acid? Solution Equation NaOH + HCl NaCl + H2O Wanted ? ml of 0.154 M NaOH Given 25.0 mL of 0.0952 M HCl Conversion factors 1 mL of 0.0952 M HCl contains 0.0952 mmol HCl 1 mol NaOH reacts with 1 mol HCl 1 mL of 0.154 M NaOH contains 0.154 mmol NaOH Arithmetic equation Answer 15.5 mL of 0.154 M NaOH

The next example illustrates the use of molarity in stoichiometric problems invloving gases.

 Example: What is the concentration of an aqueous solution of hydrochloric acid if 25.0 mL of the acid reacts with an excess of solid calcium carbonate to yield 0.307 L carbon dioxide, measured at 0.95 atm and 25°C? Solution Analysis of the problem 1. The product of the reaction is a gas whose volume was not measured at standard conditions of temperature and pressure. We can calculate the number of moles of gas produce by using the ideal gas equation: PV = nRT 2. Using the number of moles of carbon dioxide produced, we can calculate the number of moles of hydrochloric acid used. The mles of HCl are in 25.0 mL solution. Molarity is a ratio between moles of solute and volume of solution. By dividing the number of moles of HCl by the volume (L) of solution in which it was dissolved, we will obtain the molarity of the acid solution. Calculations 2. Carry out the stoichiometric calculation. equation 2 HCl + CaCO3 CO2 + CaCl2 +H2O Wanted The molarity of the acid solution or mol HCl/L solution Given 0.012 mol CO2 Conversion factor 2 mol HCl yield 1 mol CO2 Arithmetic equation Answer 0.96 M HCl