Question

A 35.0-mL sample of \(1.00 \mathrm{MCo}(\mathrm{NO})_{3}\) and an 80.0 -mL sample of \(0.600 \mathrm{M} \mathrm{Co}(\mathrm{NO})_{3}\) are mixed. The solution is then heated to evaporate water until the total volume is \(50.0 \mathrm{~mL}\). Calculate the volume, in \(\mathrm{mL}\), of \(0.20 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) that is required to precipitate out cobalt (III) phosphate in the final solution.

Short answer

415 mL of 0.20 M \( \mathrm{H}_3\mathrm{PO}_4 \) is required.

Step-by-step solution

Calculate Moles in Each Initial Mixture

First, determine how many moles of \( \mathrm{Co(NO)}_{3} \) are in each solution before mixing. Use the formula \( \text{moles} = \text{Molarity} \times \text{Volume (L)} \). For the first solution, \( \text{moles} = 1.00 \, \text{M} \times 0.035 \, \text{L} = 0.035 \, \text{mol} \). For the second solution, \( \text{moles} = 0.600 \, \text{M} \times 0.080 \, \text{L} = 0.048 \, \text{mol} \).

Calculate Total Moles After Mixing

Add the moles from each solution together to get the total moles of \( \mathrm{Co(NO)}_{3} \) after mixing. Total moles = \( 0.035 \, \text{mol} + 0.048 \, \text{mol} = 0.083 \, \text{mol} \).

Calculate Concentration in the Evaporated Solution

The total volume after evaporation is given as 50.0 mL or 0.050 L. Calculate the new concentration using the formula \( \text{Molarity} = \frac{\text{moles}}{\text{volume (L)}} \). Thus, \( \text{Molarity} = \frac{0.083 \, \text{mol}}{0.050 \, \text{L}} = 1.66 \, \text{M} \).

Write Reaction Equation for Precipitation

The reaction between cobalt(III) nitrate and phosphoric acid to form cobalt(III) phosphate is: \[ 2\mathrm{Co(NO}_3\mathrm{)_3} + 2\mathrm{H}_3\mathrm{PO}_4 \rightarrow \mathrm{Co}_2(\mathrm{PO}_4)_3 \downarrow + 6\mathrm{HNO}_3 \]. The stoichiometry shows that 2 moles of \( \mathrm{Co(NO)}_{3} \) reacts with 2 moles of \( \mathrm{H}_3\mathrm{PO}_4 \).

Calculate Moles of \( \mathrm{H}_3\mathrm{PO}_4 \) Required

According to the reaction equation, 2 moles of \( \mathrm{Co(NO)}_{3} \) require 2 moles of \( \mathrm{H}_3\mathrm{PO}_4 \). Therefore, for 0.083 mol of \( \mathrm{Co(NO)}_{3} \), 0.083 mol of \( \mathrm{H}_3\mathrm{PO}_4 \) are needed.

Calculate Volume of \( \mathrm{H}_3\mathrm{PO}_4 \) Solution

Use the formula \( \text{Volume (L)} = \frac{\text{moles}}{\text{Molarity}} \). For 0.083 mol required and a concentration of 0.20 M, the volume \( V = \frac{0.083 \, \text{mol}}{0.20 \, \text{M}} = 0.415 \, \text{L} \). Convert this to mL: 0.415 L = 415 mL.

Key concepts

Molarity

Molarity is a measure of concentration in chemistry. It shows how many moles of a solute are in one liter of solution. Calculating molarity helps us know the concentration of chemicals in a reaction. Here’s how you calculate it:

• Identify the number of moles of solute (the substance dissolved).
• Measure the total volume of the solution in liters.
• Use the formula: \( \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in L}} \)

In the problem, the total moles of cobalt nitrate are calculated first by using the initial molarity and volumes given: \( 1.00 \,\text{M} \times 0.035 \,\text{L} \) for one solution and \( 0.600 \,\text{M} \times 0.080 \,\text{L} \) for the other. After calculating the moles, you combine them to find the total moles in the new mixture. Molarity is used once more to find the new concentration after the solution volumes change, demonstrating how molarity is vital in tracking chemical concentrations throughout reactions.

Chemical reactions

Chemical reactions involve substances (reactants) transforming into different substances (products). Each reaction follows a balanced chemical equation. This equation lets us understand the stoichiometry, showing the number and types of atoms conserved. Types of reactions include combination, decomposition, and replacement. For this exercise, attention is on how cobalt nitrate reacts with phosphoric acid. The balanced equation is:\[ 2\mathrm{Co(NO}_3\mathrm{)_3} + 2\mathrm{H}_3\mathrm{PO}_4 \rightarrow \mathrm{Co}_2(\mathrm{PO}_4)_3 \downarrow + 6\mathrm{HNO}_3 \]. In this equation:

• 2 moles of cobalt nitrate react with 2 moles of phosphoric acid.
• The reactions produce cobalt phosphate (precipitate) and nitric acid.

Knowing the balanced equation is crucial for calculating each reactant needed for the reaction, as it shows the stoichiometric relationships among components.

Precipitation reactions

Precipitation reactions are when two aqueous solutions combine, forming an insoluble solid known as a precipitate. The solid drops out of the liquid mixture, visible as a cloudy suspension or a solid at the bottom. In these reactions, knowing which reactants form a precipitate is important.
In this exercise, cobalt(III) phosphate is the precipitate. The cobalt nitrate solution reacts with phosphoric acid to form it. As shown:\[ 2\mathrm{Co(NO}_3\mathrm{)_3} + 2\mathrm{H}_3\mathrm{PO}_4 \rightarrow \mathrm{Co}_2(\mathrm{PO}_4)_3 \downarrow + 6\mathrm{HNO}_3 \].
To balance this reaction, it is important to understand that each component reacts in specific ratios. This stoichiometry helps calculate how much of each reactant is necessary to produce the desired amount of precipitate. In the above example, each mole of cobalt nitrate pairs with a mole of phosphoric acid in a 1:1 ratio to fully precipitate cobalt phosphate. Given the need to achieve a certain concentration or remove specific ions from a solution, these reactions play an important role in chemistry.