Problem 74 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) How many milliliters of a stock solution of \(6.0 \mathrm{MHNO}_{3}\) would you have to use to prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{MHNO}_{3} ?\) (b) If you dilute \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L},\) what will be the concentration of the diluted solution?

Step-by-Step Solution

Answer

(a) To prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{MHNO}_{3}\) solution, you would need to use \(9.167 \mathrm{~mL}\) of the \(6.0 \mathrm{MHNO}_{3}\) stock solution. (b) After diluting \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L}\), the concentration of the diluted solution will be \(0.24 \mathrm{M}\).

Step 1: (a) Finding the volume of stock solution needed for the diluted solution

: We can use the dilution formula to solve this problem: M1V1 = M2V2 In this case, M1 = 6.0 M (stock solution concentration), M2 = 0.500 M (diluted solution concentration), and V2 = 110 mL (diluted solution volume). We want to find V1 (volume of stock solution required). Rearranging the formula for V1: V1 = M2V2 / M1 Substituting the known values: V1 = (0.500 M × 110 mL) / 6.0 M Calculating the volume: V1 = 9.167 mL Therefore, 9.167 mL of 6.0 MHNO3 stock solution would be required to prepare 110 mL of 0.500 MHNO3 solution.

Step 2: (b) Finding the concentration of the diluted solution after dilution

: Again, we'll use the dilution formula for this part: M1V1 = M2V2 In this case, M1 = 6.0 M (stock solution concentration), V1 = 10 mL (stock solution volume used), V2 = 0.250 L (final volume of the diluted solution). We want to find M2 (concentration of the diluted solution). Rearranging the formula for M2: M2 = M1V1 / V2 Note that we need to make sure that our volume units are consistent. With V2 given in L, we will convert V1 to L. V1 = 10 mL = 0.010 L Now, we can substitute the values into the formula: M2 = (6.0 M × 0.010 L) / 0.250 L Calculating the concentration: M2 = 0.24 M Therefore, the concentration of the diluted solution after using 10.0 mL of the stock solution and a final volume of 0.250 L is 0.24 M.

Key Concepts

Molar Concentration

Molar concentration, often denoted as simply "concentration," tells us how much of a substance (solute) is present per unit volume of solution. Imagine molar concentration as the number of "boxes" of solute in each liter of the solution.
It’s measured in moles per liter (M), where one mole is approximately 6.022 x 10²³ entities, like atoms or molecules. This measurement helps chemists understand the strength or "intensity" of a solution's solute.
For example, in this exercise, we calculate molar concentration using the formula:

• Molarity (M) = Moles of solute / Volume of solution in liters

This formula allows us to determine how concentrated or diluted a solution is, which is critical for tasks like chemical reactions or creating solutions with specific properties.

Stock Solution

A stock solution is a concentrated solution that can be diluted to a lower concentration for actual use. Imagine it as a concentrated beverage from which you make drinks with varying strengths by adding water.
Stock solutions are used to save space and resources, as you can store higher concentrations more conveniently. You only need a small amount of this concentrated stock to prepare a new solution with your desired concentration.
In the example problem, we have a stock solution of nitric acid (\( \mathrm{HNO}_3 \)) with a concentration of 6.0 M. To achieve a specific concentration in a new solution, we simply calculate how much of this stock needs to be mixed with solvent, usually water.

Solution Preparation

Solution preparation involves mixing the right proportions of solute and solvent to achieve a desired concentration. This is like preparing a recipe, where exact amounts are mixed to achieve the perfect flavor.
A common formula used in dilution calculations is:

• M₁V₁ = M₂V₂

Here, M₁ and V₁ represent molarity and volume of the stock solution, whereas M₂ and V₂ are the molarity and volume of the diluted solution.
This equation illustrates the conservation of moles between the concentrated and diluted solutions. By rearranging this equation, you can compute how much of the stock solution is needed to create the desired final solution.

Diluted Solution

A diluted solution is simply one that has been reduced from its original concentration by adding more solvent, thereby decreasing the solute concentration.
This is like adding more water to orange juice to make it less strong. In chemistry, diluting solutions is crucial for obtaining the right concentration needed for experiments.
For the exercise at hand, we calculated what happens when a stock solution is diluted. We used the formula:

• M₂ = (M₁V₁) / V₂

When we know the initial concentration and the volumes, we pinpoint the new concentration of our diluted solution. This concept ensures precision in scientific experiments, helping achieve consistent and reliable results.