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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.

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 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.