Calculations Involving Precipitation Reactions Suppose we want to prepare BaSO4 by adding 0.450 M K2SO4 to 130.0 mL of 0.250 M BaCl2. What volume of the K2SO4 solution is required to react completely with the BaCl2? How many grams of BaSO4 will precipitate?

Question

Calculations Involving Precipitation Reactions

Suppose we want to prepare BaSO₄ by adding 0.450 M K₂SO₄ to 130.0 mL of 0.250 M BaCl₂. What volume of the K₂SO₄ solution is required to react completely with the BaCl₂? How many grams of BaSO₄ will precipitate?

Accepted Answer

For the first part of this problem we want to go from the volume of one reactant to the volume of another:

[latex]\boxed{ ext{Volume} BaCl_{2} ext{solution}}\overset{?}{\longrightarrow } \boxed{ ext{Volume} K_{2}SO_{4} ext{solution}}[/latex]

There is no direct relationship between these two quantities, but this is a problem in which we’re given a quantity of one substance in a chemical reaction and want to calculate another. We’ll have to refer to the balanced equation to determine the mole relationship for the two substances. From the balanced equation we know that 1 mol of BaCl₂ reacts with 1 mol of K₂SO₄ to produce 1 mol of BaSO₄. Is there a relationship between the volume of BaCl₂ and moles of BaCl₂?

[latex]\boxed{ ext{Volume} BaCl_{2} ext{solution}}\overset{?}{\longrightarrow } \boxed{mol BaCl_{2} }[/latex]

Yes, its molarity is 0.250 M, which tells us that for every liter of solution, 0.250 mol of BaCl₂ is present. In Chapter 6 we obtained moles of reactant from grams using the molar mass as a conversion factor. Here we must convert the volume of a reactant in solution to moles using the molar concentration as a conversion factor. In this case we know the volume of 0.250 M BaCl₂ is 130.0 mL. We have to first convert milliliters to liters because the volume unit in molarity is liters:

[latex]\boxed{ ext{Volume} BaCl_{2} (in mL)}\overset{1000 mL = 1 L}{\longrightarrow } \boxed{Volume BaCl_{2} (in L)}\overset{molarity}{\longrightarrow } \boxed{mol BaCl_{2} }[/latex]

[latex]mol BaCl_{2} = 130.0 \cancel{mL BaCl_{2} ext{solution}} imes \frac{1 L}{1000 \cancel{mL}} imes \frac{0.250 mol BaCl_{2} }{\cancel{L BaCl_{2} ext{solution}} } = 0.0325 mol BaCl_{2}[/latex]

We can now use the mole ratio in the chemical equation to calculate the moles of K₂SO₄ required for the reaction:

[latex]\boxed{mol BaCl_{2} }\overset{ ext{mole ratio}}{\longrightarrow } \boxed{mol K_{2}SO_{4}}[/latex]

[latex]mol K_{2}SO_{4} = 0.0325 \cancel{mol BaCl_{2} } imes \frac{1 mol K_{2}SO_{4} }{\cancel{mol BaCl_{2} }}= 0.0325 mol K_{2}SO_{4}[/latex]

Now that we know the moles of K₂SO₄ required to completely react with the BaCl₂, we have to determine the volume of K₂SO₄ solution that will provide the desired number of moles:

[latex]\boxed{mol K_{2}SO_{4} }\overset{?}{\longrightarrow } \boxed{ ext{Volume} K_{2}SO_{4} ext{solution (in L)}}[/latex]

The molar concentration of K₂SO₄ solution provides a relationship we can use to calculate the volume:

[latex]\boxed{mol K_{2}SO_{4} }\overset{molarity}{\longrightarrow } \boxed{ ext{Volume} K_{2}SO_{4} ext{solution (in L)}}[/latex]

[latex] ext{Volume of} K_{2}SO_{4} ext{solution} = 0.0325 \cancel{mo l K_{2}SO_{4} } imes \frac{1 L K_{2}SO_{4} ext{solution} }{0.450 \cancel{mol K_{2}SO_{4} }} = 0.0722 L K_{2}SO_{4} ext{solution} [/latex]

We can interpret this volume of K₂SO₄ solution as the amount of solution that has enough moles of K₂SO₄ (0.0325 mol) to react with all the BaCl₂. An alternative to this step-by-step approach involves setting up conversions consecutively in one calculation. To determine the volume of K₂SO₄ required, we can set up the following sequence of conversions:

[latex] ext{Volume of } K_{2}SO_{4} ext{solution} = 0.1300 \cancel{L BaCl_{2} ext{solution}} imes \frac{0.250 \cancel{mol BaCl_{2} }}{1 \cancel{L BaCl_{2} ext{solution}}} imes \frac{1 \cancel{mol K_{2}SO_{4} }}{1 \cancel{ BaCl_{2} }} imes \frac{1 L K_{2}SO_{4} ext{solution}}{0.450 \cancel{mol K_{2}SO_{4} }}= 0.0722 L K_{2}SO_{4} ext{solution}[/latex]

This is a small number more conveniently expressed in milliliters:

[latex] ext{Volume in mL} = 0.0722 \cancel{L} K_{2}SO_{4} ext{solution} imes \frac{1000 mL}{1 \cancel{L} } = 72.2 mL K_{2}SO_{4} ext{solution}[/latex]

The following sequence summarizes the relationships used to solve this problem:

[latex]\boxed{ ext{Volume (in L)} BaCl_{2} ext{solution}}\overset{molarity}{\longrightarrow } \boxed{mol BaCl_{2} }\overset{mole ratio}{\longrightarrow } \boxed{mol K_{2}SO_{4} }\overset{molarity}{\longrightarrow } \boxed{ ext{Volume (in L)} K_{2}SO_{4} ext{solution}} [/latex]

Now let’s calculate the mass of solid BaSO₄ that will be produced when adding 72.2 mL of 0.450 M K₂SO₄ solution to 130.0 mL of 0.250 M BaCl₂. We know the moles of BaCl₂ from a previous calculation. We can use the mole ratio given by the balanced chemical equation to determine the moles of BaSO₄ that should form:

[latex]mol BaSO_{4} = 0.0325 \cancel{mol BaCl_{2} } imes \frac{1 mol BaSO_{4} }{1 \cancel{mol BaCl_{2} }}= 0.0325 mol BaSO_{4}[/latex]

Do you recall how to calculate grams of a substance from moles of that material? Using the molar mass, we can calculate the grams of BaSO₄ that will precipitate:

[latex]\boxed{mol BaSO_{4} }\overset{molar mass}{\longrightarrow } \boxed{Mass BaSO_{4}}[/latex]

[latex]Mass of BaSO_{4}= 0.0325 \cancel{mol BaSO_{4}} imes \frac{233.4 g BaSO_{4}}{1 \cancel{mol BaSO_{4}}}= 7.59 g BaSO_{4}[/latex]

An alternative to this step-by-step approach involves setting up conversions consecutively in one calculation. To determine the mass of BaSO₄ produced, we can set up the following sequence of conversions:

[latex]Mass of BaSO_{4}= 0.1300 \cancel{L BaCl_{2} ext{solution} } imes \frac{0.250 \cancel{mol BaCl_{2} }}{1 \cancel{ L BaCl_{2} ext{solution}}} imes \frac{1 \cancel{mol BaSO_{4} }}{1 \cancel{mol BaCl_{2} }} imes \frac{233.4 g BaSO_{4} }{1 \cancel{mol BaSO_{4} } } = 7.59 g BaSO_{4} [/latex]

[latex]\boxed{ ext{Volume (in L)} BaCl_{2} }\overset{molarity}{\longrightarrow } \boxed{mol BaCl_{2}} \overset{mole ratio}{\longrightarrow } \boxed{mol BaSO_{4}} \overset{molar mass}{\longrightarrow } \boxed{mass BaSO_{4}}[/latex]

The key to solving problems that involve solution concentrations is to realize that concentration is a ratio that can be used as a conversion factor. If volume is known and we want to calculate moles, we simply multiply by the molarity. If moles are given and we want to calculate volume, we divide by molarity.

Question 11.8: Calculations Involving Precipitation Reactions Suppose we wa...

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