Question

You have two solutions, one \(0.755 \mathrm{M}\) barium nitrate and the other \(1.250 \mathrm{M}\) calcium hydroxide. (a) Write a net ionic equation for the precipitation reaction that occurs when these solutions are combined. (b) How many milliliters of the two solutions must be combined to prepare \(5.00 \mathrm{~g}\) of precipitate? (c) Suppose you filter off the precipitate and find that your percent yield is \(85.0 \%\). What volumes of the solutions should you have combined to isolate \(5.00 \mathrm{~g}\) of precipitate?

Short answer

(a) The net ionic equation for the precipitation reaction between barium nitrate and calcium hydroxide is: \([\text {Ba}^{2+}]_{(aq)} + 2[\text {OH}^{-}]_{(aq)} \rightarrow \text {Ba(OH)}_2 (s)\) (b) To prepare 5.00 g of precipitate, combine 39.08 mL of 0.755 M barium nitrate solution and 47.26 mL of 1.25 M calcium hydroxide solution. (c) To isolate 5.00 g of precipitate with an 85% yield, combine 45.98 mL of 0.755 M barium nitrate solution and 55.60 mL of 1.25 M calcium hydroxide solution.

Step-by-step solution

1. Identify the formula and solubility of the compounds

First, let's identify the formula of each compound: Barium nitrate: \(Ba(NO_3)_2\) Calcium hydroxide: \(Ca(OH)_2\) The formation of precipitate occurs when two soluble ions combine to form an insoluble compound. In this case, the precipitate is barium hydroxide (\(Ba(OH)_2\)).

2. Write the balanced molecular equation

The balanced molecular equation for the reaction between barium nitrate and calcium hydroxide is: \[\text {Ba(NO}_3)_2 (aq) + \text {Ca(OH)}_2 (aq) \rightarrow \text {Ba(OH)}_2 (s) + \text {Ca(NO}_3)_2 (aq)\]

3. Write the net ionic equation

First, we will separate the soluble compounds into their individual ions: \([\text {Ba}^{2+}]_{(aq)} + 2[\text {NO}_3 ^{-}]_{(aq)} \) + \([\text {Ca}^{2+}]_{(aq)} + 2[\text {OH}^{-}]_{(aq)}\) \(\rightarrow [\text {Ba}^{2+}]_{(aq)} + 2[\text {OH}^{-}]_{(aq)} \rightarrow \text {Ba(OH)}_2 (s)\) \(+ [\text {Ca}^{2+}]_{(aq)} + 2[\text {NO}_3 ^{-}]_{(aq)}\) The net ionic equation for this reaction (omitting spectator ions) is: \([\text {Ba}^{2+}]_{(aq)} + 2[\text {OH}^{-}]_{(aq)} \rightarrow \text {Ba(OH)}_2 (s)\)

4. Calculating the volumes of the two solutions

To find the volumes needed to prepare 5.00 grams of barium hydroxide, we will be using stoichiometry and molarity relationship. First, convert the mass (5.00 grams) of barium hydroxide into moles using its molar mass (137.33 + 2 * 15.999 = 169.328 g/mol): Moles of Barium hydroxide = \(\frac{5.00 \text { g}}{169.328 \frac{\text {g}}{\text{mol}}} = 0.02953 \text { mol}\) According to the stoichiometry of the net ionic equation, one mole of barium hydroxide forms from one mole of barium ions: Moles of Barium ions = \(0.02953 \text { mol Ba}^{2+}\) Now, we can find the volume of 0.755 M barium nitrate solution needed: Volume of barium nitrate = \(\frac{0.02953 \text { mol Ba}^{2+}}{0.755 \frac{\text {moles}}{\text {L}}} = 0.03908 \mathrm{L} = 39.08 \mathrm{mL}\) The stoichiometry of the net ionic equation also shows that one mole of barium hydroxide forms from two moles of hydroxide ions: Moles of hydroxide ions = \(2 \times 0.02953 \text { mol OH}^{-} = 0.05907 \text { mol}\) Now, we can find the volume of 1.25 M calcium hydroxide solution needed: Volume of calcium hydroxide = \(\frac{0.05907 \text { mol OH}^{-}}{1.25 \frac{\text {moles}}{\text {L}}} = 0.047256 \mathrm{L} = 47.26 \mathrm{mL}\)

5. Adjusting for the 85% yield

Now, we have to adjust the volumes of both solutions to account for the 85% yield. To do so, we will divide the calculated volumes by the percent yield: Adjusted volume of barium nitrate = \(\frac{39.08 \text { mL}}{0.85} = 45.98 \text {mL}\) Adjusted volume of calcium hydroxide = \(\frac{47.26 \text { mL}}{0.85} = 55.60 \text {mL}\) Thus, to isolate 5.00 g of precipitate with an 85% yield, 45.98 mL of 0.755 M barium nitrate solution and 55.60 mL of 1.25 M calcium hydroxide solution should be combined.

Key concepts

Net Ionic Equation

Understanding the net ionic equation is crucial when studying chemical reactions, especially precipitation reactions. The net ionic equation distills a chemical reaction down to its most fundamental form, showing only the species involved in the actual reaction and omitting the spectator ions that do not participate.

For a precipitation reaction, like the one between barium nitrate and calcium hydroxide, you begin by writing the full ionic equation, which includes all aqueous substances as ions. Then, you identify the precipitate, the solid substance formed during the reaction, which in this case is barium hydroxide (Ba(OH)₂). Spectator ions, such as nitrate (NO₃^-) and calcium (Ca²⁺), which remain unchanged and in solution, are omitted to reveal the net ionic equation.

In this example:\([\text{Ba}^{2+}]_{(aq)} + 2[\text{OH}^{-}]_{(aq)} \rightarrow \text{Ba(OH)}_2 (s)\), barium ions (Ba²⁺) react with hydroxide ions (OH^-) to form the insoluble barium hydroxide. Understanding this concept is pivotal for predicting reaction outcomes and grasping the essentials of inorganic chemistry.

Stoichiometry

Stoichiometry is the section of chemistry that pertains to the calculation of reactants and products in chemical reactions. It is based on the law of conservation of mass, where the amount of each element must be the same before and after a chemical reaction.

In stoichiometry, coefficients of a balanced chemical equation are interpreted as the relative amounts (in moles) of each substance involved. For instance, when calculating the volumes necessary to create a certain amount of precipitate, you would use stoichiometry to convert between mass, moles, and volume using molar masses and molar concentration.

The stoichiometry of our reaction indicates that each mole of barium hydroxide is formed from one mole of barium ion and two moles of hydroxide ions. This stoichiometric relationship allows us to find the precise volumes of the starting solutions needed to create a desired amount of precipitate, ensuring that the proportions of reactants are correct to optimize yield and minimize waste.

Molar Concentration

Molar concentration, often referred to as molarity and expressed as moles per liter (M), is a measure of the concentration of a solute in a solution. It's an invaluable concept for chemistry students, as it quantifies how much of a substance is present in a given volume of liquid.

Molarity plays a key role in preparing solutions and conducting reactions where specific amounts of reactants are needed. In our exercise, we determined the volume of a 0.755 M barium nitrate solution and a 1.25 M calcium hydroxide solution required to produce 5.00 grams of barium hydroxide precipitate.

By converting the desired mass of precipitate to moles and then using the molar concentrations of the solutions, we can calculate the exact volumes needed for the reaction. For example, to find the volume of the barium nitrate solution, divide the number of moles of barium ion by the molarity of the solution. In teaching and practice, understanding molar concentration simplifies complex chemical calculations and enables precise control over the quantities of reagents in the laboratory.