Problem 104 - Chemistry: An Atoms-Focused Approach

Question

The solubility of magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2},\) in water is \(9.0 \times 10^{-4} \mathrm{g} / 100.0 \mathrm{mL} .\) What volume of \(0.00100 M \mathrm{HNO}_{3}\) is required to neutralize \(1.00 \mathrm{L}\) of a saturated Mg(OH), solution?

Step-by-Step Solution

Answer

Answer: The volume of 0.00100 M HNO₃ required to neutralize 1.00 L of a saturated Mg(OH)₂ solution is 0.3084 L.

Step 1: Write down the balanced chemical equation for the neutralization reaction

The balanced equation for the reaction between magnesium hydroxide and nitric acid is: \(\mathrm{Mg}(\mathrm{OH})_{2}(s) + 2\mathrm{HNO}_{3}(aq) \rightarrow \mathrm{Mg}(\mathrm{NO}_{3})_{2}(aq) + 2\mathrm{H}_{2}\mathrm{O}(l)\)

Step 2: Calculate the moles of magnesium hydroxide in the solution.

Given the solubility of magnesium hydroxide as \(9.0 \times 10^{-4} \mathrm{g} / 100.0 \mathrm{mL}\), we can calculate the moles of magnesium hydroxide in the 1.00 L saturated solution. First, convert solubility from mg/100 mL to g/L: Solubility of Mg(OH)₂ \(= 9.0 \times 10^{-4} \mathrm{g} / 100.0 \mathrm{mL} \times \frac{1000 \mathrm{mL}}{1 \mathrm{L}} = 9.0 \times 10^{-3} \mathrm{g/L}\) Next, calculate the moles of Mg(OH)₂ using its molar mass (58.32 g/mol): Moles of Mg(OH)₂ \(= \frac{9.0 \times 10^{-3} \mathrm{g}}{58.32 \mathrm{g/mol}} = 1.542 \times 10^{-4} \mathrm{mol}\)

Step 3: Determine the moles of nitric acid required for neutralization.

From the balanced chemical equation, one mole of Mg(OH)₂ reacts with two moles of HNO₃. Therefore, we can determine the moles of HNO₃ needed for neutralization: Moles of HNO₃ = 2 × moles of Mg(OH)₂ Moles of HNO₃ = 2 × 1.542 × 10⁻⁴ mol = 3.084 × 10⁻⁴ mol

Step 4: Calculate the volume of HNO₃ solution required.

We are given the concentration of HNO₃ as 0.00100 M (moles per liter). The volume of HNO₃ solution required for the neutralization can be calculated as: Volume of HNO₃ = \(\frac{\text{moles of HNO₃}}{\text{concentration of HNO₃}}\) Volume of HNO₃ = \(\frac{3.084 \times 10^{-4} \ \mathrm{mol}}{0.00100 \ \mathrm{M}} = 0.3084 \ \mathrm{L}\) So, the volume of 0.00100 M HNO₃ required to neutralize 1.00 L of a saturated Mg(OH)₂ solution is 0.3084 L.

Key Concepts

Solubility

Solubility is the ability of a substance to dissolve in a solvent, forming a solution at a certain temperature and pressure. It tells us the maximum amount of solute that can be dissolved in a solvent to form a homogeneous mixture, or solution. In the exercise, the solubility of magnesium hydroxide, \(\mathrm{Mg(OH)}_{2}\), is provided as

• \(9.0 \times 10^{-4}\, \mathrm{g} / 100.0\, \mathrm{mL}\).

This means only a small amount of \(\mathrm{Mg(OH)}_{2}\) can dissolve in water. To use this information for calculations, we need to express solubility in a common measurement, such as grams per liter (g/L). This is achieved by converting the unit to g/L:

• \(9.0 \times 10^{-4} \times \frac{1000 \mathrm{mL}}{100 \mathrm{mL}} = 9.0 \times 10^{-3}\, \mathrm{g/L}\).

Knowing the solubility helps determine how much of \(\mathrm{Mg(OH)}_{2}\) is available to react in a solution of a given volume.

Moles Calculation

Calculating moles is a fundamental skill in chemistry, important for understanding how much of a substance is involved in a reaction. The number of moles can be calculated using the formula:

• \(\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\).

For magnesium hydroxide in this problem, the molar mass is crucial. The molar mass of \(\mathrm{Mg(OH)}_{2}\) is 58.32 g/mol. Given the converted solubility:

• \(9.0 \times 10^{-3}\, \mathrm{g/L}\),

we can find the moles in 1.00 L of solution:

• Moles of \(\mathrm{Mg(OH)}_{2} = \frac{9.0 \times 10^{-3} \, \mathrm{g}}{58.32 \, \mathrm{g/mol}} = 1.542 \times 10^{-4} \, \text{mol}\).

This calculation tells us how much \(\mathrm{Mg(OH)}_{2}\) is present to participate in further chemical reactions, like neutralization.

Molarity

Molarity is the concentration measure of a solute in a solution. It is expressed as the number of moles of a solute per liter of solution and calculated using:

• \(\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\).

In our exercise, we have a \(0.00100 \, \mathrm{M}\) \(\mathrm{HNO}_{3}\) solution, which refers to the molarity. To find the volume of \(\mathrm{HNO}_{3}\) required to neutralize the \(\mathrm{Mg(OH)}_{2}\), we used the relationship:

• \(\text{Volume of } \mathrm{HNO}_{3} = \frac{3.084 \times 10^{-4} \, \mathrm{mol}}{0.00100 \, \mathrm{M}}\).

This division gives the volume of \(\mathrm{HNO}_{3}\) required to provide the needed moles to react completely with the moles of \(\mathrm{Mg(OH)}_{2}\) in the solution. Understanding molarity helps in making solutions and carrying out accurate neutralization reactions in chemistry.