Question

To prepare a very dilute solution, it is more accurate to make up a more concentrated standard solution, and carry out a series of successive dilutions, than to weigh out a very small mass of the solute. A solution was made by dissolving \(0.587 \mathrm{g}\) of \(\mathrm{KMnO}_{4}\) in dilute sulfuric acid and making the volume of solution up to \(1 \mathrm{dm}^{3}\) in a volumetric flask. \(10.0 \mathrm{cm}^{3}\) of this solution were transferred to a second \(1 \mathrm{dm}^{3}\) volumetric flask and diluted to the mark with water. The dilution process was then repeated once, that is, \(10.0 \mathrm{cm}^{3}\) of this solution were transferred to a \(1 \mathrm{dm}^{3}\) volumetric flask and diluted to the mark with water. (Section 1.5 ). (a) What mass (in \(\mathrm{mg}\) ) of \(\mathrm{KMnO}_{4}\) would you have had to weigh out to make \(500 \mathrm{cm}^{3}\) of a solution with the same concentration as the final dilute solution? (b) What is the concentration of the final dilute \(\mathrm{KMnO}_{4}\) solution in moldm \(^{-3} ?\)

Short answer

(a) 0.0293 mg, (b) \(3.71 \times 10^{-7}\) mol/dm³.

Step-by-step solution

Calculate the initial concentration

The initial solution was made by dissolving 0.587 g of \( \mathrm{KMnO}_4 \) in 1 dm³ of solution. The molar mass of \( \mathrm{KMnO}_4 \) is about 158.04 g/mol. The concentration \( C_1 \) is given by \( C_1 = \frac{\text{mass}}{\text{molar mass} \times \text{volume in dm³}} = \frac{0.587}{158.04 \times 1} \approx 0.00371 \; \text{mol dm}^{-3} \).

First dilution concentration

\( 10.0 \; \mathrm{cm}^3 \) (or \( 0.010 \; \mathrm{dm}^3 \)) of the initial solution is diluted to 1 dm³. Using the dilution formula \( C_2 = \frac{C_1 \times V_1}{V_2} \), where \( V_1 = 0.010 \; \text{dm}^3 \) and \( V_2 = 1 \; \text{dm}^3 \), results in \( C_2 = \frac{0.00371 \times 0.010}{1} = 0.0000371 \; \text{mol dm}^{-3} \).

Second dilution concentration

Again, \( 10.0 \; \mathrm{cm}^3 \) of \( C_2 \) is diluted to 1 dm³. The concentration \( C_3 \) becomes \( C_3 = \frac{C_2 \times 0.010}{1} = \frac{0.0000371 \times 0.010}{1} = 0.000000371 \; \text{mol dm}^{-3} \).

Calculate mass for equivalent concentration

The concentration of the final dilution is \( 0.000000371 \; \mathrm{mol\,dm}^{-3} \). To prepare 500 cm³ (0.500 dm³) of this solution, the mass needed \( m \) is calculated using \( m = C_3 \times \text{molar mass} \times \text{volume in dm}^3 \). Thus, \( m = 0.000000371 \times 158.04 \times 0.500 = 0.0000293 \; \text{g} = 0.0293 \; \text{mg} \).

Final concentration in mol/dm³

Since we already calculated \( C_3 = 0.000000371 \; \text{mol dm}^{-3} \), this is the concentration in moldm \(^{-3}\) of the final solution.

Key concepts

Concentration Calculation

Understanding concentration calculation is crucial when working with solutions. Concentration is often expressed in moles per cubic decimeter (moldm\(^{-3}\)), which tells us how much solute is dissolved in a given volume of solvent. To compute concentration, you'll need:

• The mass of the solute.
• The molar mass of the solute.
• The total volume of the solution.

A simple formula for concentration is: \[C = \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{volume of solution}}\]This formula helps determine how strong or weak a solution is based on the amount of solute present. It gives us a precise way to express concentration, which is especially useful in chemical reactions and quality control.

Volumetric Flask

A volumetric flask is an essential tool used in chemistry labs for preparing solutions with precise volumes. These flasks are designed to hold a specific volume and usually have a narrow neck with a calibration mark. Here's why volumetric flasks are important:

• They allow for accurate measurement of liquid volumes, which is vital when preparing solutions with specific concentrations.
• They are ideal for making dilutions precisely, as the flask's narrow neck helps achieve this.
• Each flask is calibrated for a specific volume, ensuring consistent results across different experiments.

When using a volumetric flask, it's crucial to add the liquid up to the mark gently to avoid overshooting the required volume. The precise design makes these flasks indispensable in any laboratory setting where accuracy is needed.

Dilution Formula

The dilution formula is a handy tool for adjusting the concentration of a solution. It accounts for changes in concentration when a solution is diluted. The formula is: \[C_1 \times V_1 = C_2 \times V_2\]Where:

• \(C_1\) is the initial concentration.
• \(V_1\) is the initial volume.
• \(C_2\) is the final concentration.
• \(V_2\) is the final volume.

This equation helps predict the concentration change after diluting a solution. When you dilute, the concentration decreases, as you have more solvent compared to solute. Understanding this concept is crucial for calculating the final concentration of a solution after multiple dilution steps, ensuring precise control over chemical reactions.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Calculating molar mass involves summing the atomic masses of each element in a compound. For instance, the molar mass of potassium permanganate (\(\text{KMnO}_4\)) is determined by adding:

• Atomic mass of K (Potassium): approximately 39.10 g/mol.
• Atomic mass of Mn (Manganese): approximately 54.94 g/mol.
• Atomic mass of O (Oxygen): approximately 16.00 g/mol each, and there are four oxygens: 4 × 16.00 = 64.00 g/mol.

The total molar mass of \(\text{KMnO}_4\) is approximately 158.04 g/mol. Knowing how to calculate molar mass is vital for converting between grams and moles, which is commonly required in stoichiometry and concentration calculations.