Question

Consider a \(13.0 \%\) solution of sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), whose density is \(1.090 \mathrm{~g} / \mathrm{mL}\). (a) Calculate the molarity of this solution. (b) To what volume should \(100 . \mathrm{mL}\) of this solution be diluted to prepare a 1.10 -M solution?

Short answer

(a) The molarity of the solution is 1.445 M. (b) Dilute to 131.36 mL.

Step-by-step solution

Calculate the Mass of Solution in 1 Liter

First, we need to calculate the mass of 1 liter (1000 mL) of the sulfuric acid solution using its given density of \(1.090 \text{ g/mL}\). \[\text{Mass of solution} = \text{Density} \times \text{Volume} = 1.090 \text{ g/mL} \times 1000 \text{ mL} = 1090 \text{ g}\]

Find the Mass of H2SO4 in the Solution

The solution contains \(13.0\%\) sulfuric acid by mass. Calculate the mass of \(\mathrm{H}_2\mathrm{SO}_4\) in 1090 g of solution.\[\text{Mass of } \mathrm{H}_2\mathrm{SO}_4 = \frac{13.0}{100} \times 1090 \text{ g} = 141.7 \text{ g}\]

Calculate Moles of H2SO4

Now, calculate the number of moles of \(\mathrm{H}_2\mathrm{SO}_4\) in 141.7 g. Use the molar mass of \(\mathrm{H}_2\mathrm{SO}_4\), which is \(98.08 \text{ g/mol}\).\[\text{Moles of } \mathrm{H}_2\mathrm{SO}_4 = \frac{141.7 \text{ g}}{98.08 \text{ g/mol}} \approx 1.445 \text{ moles}\]

Calculate Molarity of the Solution

The molarity of a solution is defined as the moles of solute per liter of solution. From Step 3, we have \(1.445\) moles of \(\mathrm{H}_2\mathrm{SO}_4\) in 1 liter of solution.\[\text{Molarity} = \frac{\text{Moles of } \mathrm{H}_2\mathrm{SO}_4}{\text{Volume in liters}} = \frac{1.445 \text{ moles}}{1 \text{ L}} = 1.445 \text{ M}\]

Use the Dilution Formula

To find the volume needed to prepare a \(1.10\) M solution from a \(1.445\) M solution, use the dilution formula \(M_1 V_1 = M_2 V_2\). Here, \(M_1 = 1.445\) M, \(V_1 = 100 \text{ mL}\), and \(M_2 = 1.10\) M.\[1.445 \times 100 = 1.10 \times V_2\]

Solve for the Final Volume

Rearrange the equation and solve for \(V_2\).\[V_2 = \frac{1.445 \times 100}{1.10} \approx 131.36 \text{ mL}\]Thus, the solution should be diluted to \(131.36 \text{ mL}\) to achieve the desired molarity.

Key concepts

Concentration calculations

Concentration calculations are essential in chemistry as they describe the amount of solute present in a given volume of solvent. One of the most common concentration units is molarity, denoted as \(M\). Molarity is defined as the number of moles of solute per liter of solution. Calculating the molarity requires knowing the mass of the solute, its molar mass, and the volume of the solution.

To find molarity, follow these steps:

• Determine the mass of the solute in the solution. This can be done using percentage concentration by mass and the total mass of the solution.
• Calculate the number of moles of the solute by dividing its mass by its molar mass. The molar mass is a specific value with units of grams per mole for each substance.
• Finally, divide the number of moles by the volume of the solution in liters to find its molarity.

Understanding these calculations can help solve problems involving the concentration of solutions and how they change upon dilution or mixing with other solutions.

Dilution formula

The dilution formula, \(M_1V_1 = M_2V_2\), is a handy tool when you need to adjust the concentration of a solution. It's used to determine how much water or another solvent must be added to achieve a desired lower concentration.

This formula requires four variables: the initial molarity \(M_1\), the initial volume \(V_1\), the final molarity \(M_2\), and the final volume \(V_2\). Using these values, you can rearrange the formula to solve for unknowns. For example, knowing the initial concentration and amount, along with the desired concentration, allows for calculating the new total volume needed.

• Rearrange the formula to solve for the unknown.
• Use consistent units, typically moles for concentration and liters for volume.
• Understand that diluting a solution reduces its concentration but keeps the total amount of solute constant.

Mastering the dilution formula is crucial for preparing solutions in laboratory settings and ensures accurate chemical reactions.

Sulfuric acid

Sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)) is a strong mineral acid renowned for its industrial importance and key role in various chemical reactions. Its importance lies in the fact that it is widely used in the production of fertilizers, in petroleum refining, and in the processing of metals.

It has a high affinity for water and demonstrates exothermic interactions with water, indicating that mixing concentrated sulfuric acid with water releases a significant amount of heat. Therefore, it's important to carefully dilute sulfuric acid by adding acid to water, not the other way around, to safely control the heat release.

• Handle with care due to its corrosive nature.
• Used in batteries, such as car batteries, due to its acidic and conductive properties.
• Often involved in synthesis reactions as both a reagent and a catalyst.

Understanding the properties and uses of sulfuric acid is essential for anyone working with it in any chemical context.