Problem 73 - Chemistry The Central Science

Question

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{NH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

Step-by-Step Solution

Answer

(a) To make \(1000.0\mathrm{mL}\) of \(0.250\mathrm{M} \mathrm{NH}_{3}\), dilute \(16.8919\mathrm{mL}\) of the \(14.8\mathrm{M}\) stock solution. (b) The concentration of the final solution after diluting a \(10.0\mathrm{mL}\) portion of the stock solution to a total volume of \(0.500\mathrm{L}\) is \(0.296\mathrm{M}\).

Step 1: a) Calculation of milliliters of stock solution required for dilution

Using the dilution formula (\(C_1V_1 = C_2V_2\)), we can plug in the known values for initial concentration (\(C_1 = 14.8\mathrm{M}\)), final concentration (\(C_2 = 0.250\mathrm{M}\)), and final volume (\(V_2 = 1000.0\mathrm{mL}\)). Then, we solve for initial volume (\(V_1\)): \(14.8\mathrm{M} \times V_1 = 0.250\mathrm{M} \times 1000.0\mathrm{mL}\) To find the value of \(V_1\), we divide both sides of the equation by \(14.8\mathrm{M}\): \(V_1 = \frac{0.250\mathrm{M} \times 1000.0\mathrm{mL}}{14.8\mathrm{M}}\) Now, we can calculate the value of \(V_1\): \(V_1 = \frac{0.250 \times 1000.0}{14.8} = 16.8919\mathrm{mL}\) Therefore, we need to dilute \(16.8919\mathrm{mL}\) of the stock solution to make \(1000.0\mathrm{mL}\) of \(0.250\mathrm{M} \mathrm{NH}_{3}\).

Step 2: b) Calculation of the concentration of the final solution

For part (b), we have the initial concentration (\(C_1 = 14.8\mathrm{M}\)), initial volume (\(V_1 = 10.0\mathrm{mL}\)), and the final volume (\(V_2 = 0.500\mathrm{L} = 500.0\mathrm{mL}\)). We can plug these values into the dilution formula to find the final concentration (\(C_2\)): \(14.8\mathrm{M} \times 10.0\mathrm{mL} = C_2 \times 500.0\mathrm{mL}\) To solve for \(C_2\), we divide both sides of the equation by \(500.0\mathrm{mL}\): \(C_2 = \frac{14.8\mathrm{M} \times 10.0\mathrm{mL}}{500.0\mathrm{mL}}\) Now, we can calculate the value of \(C_2\): \(C_2 = \frac{14.8 \times 10.0}{500.0} = 0.296\mathrm{M}\). Therefore, the concentration of the final solution after dilution is \(0.296\mathrm{M}\).

Key Concepts

Molarity Calculations

Molarity, often represented by the symbol \(M\), is a measure of concentration used in chemistry that can tell us how much solute is present in a given volume of solution. It's defined mathematically as the number of moles of solute per liter of solution. This means if you dissolve one mole of substance in one liter of liquid, you have made a 1 Molar (1 M) solution.
Add more solute or decrease the volume, and the molarity increases. Conversely, add more solvent or increase the volume, and the molarity goes down. Understanding these interactions can be crucial in preparing solutions for experiments.\[\text{Molarity}(M) = \frac{\text{Moles of Solute}}{\text{Liters of Solution}}\]
In the given exercise, the molarity calculation is essential for both parts (a) and (b). We start with known values for molarity and use these to determine either the volume needed or the final concentration after dilution.

Dilution Formula

The dilution formula, \(C_1V_1 = C_2V_2\), is a simple yet powerful tool used to determine how much of a concentrated solution is required to achieve a desired concentration after dilution, or to find the concentration of a solution after it has been diluted.

• \(C_1\) = Initial concentration (molarity) of the solution
• \(V_1\) = Volume of the initial concentrated solution needed
• \(C_2\) = Final concentration (molarity) after dilution
• \(V_2\) = Final total volume after dilution

In our specific problem, - Part (a) involved finding \(V_1\), the volume of the concentrated solution needed to reach a 0.250 M with a total volume of 1000.0 mL. This is solved by rearranging and plugging known values into the formula. - Part (b) leveraged the same formula but was used to find the final concentration \(C_2\) after diluting a known initial volume and concentration. This flexibility makes the dilution formula an invaluable part of your chemistry toolkit.

Concentration Conversion

Concentration conversion is the process of changing the expression of a concentration to another equivalent form, often involving parts like milliliters to liters or vice versa. These conversions are particularly necessary when working with the dilution formula or molarity concepts since they demand consistent units for straightforward calculations.

In exercises like ours, you might need to convert volumes from liters to milliliters or from milliliters to liters depending on the values given. Remember,

• 1 liter = 1000 milliliters
• Ensure all units are harmonized, particularly when using the dilution formula, to prevent any mathematical errors.

By consistently checking and converting units to something manageable, especially milliliters to liters for final volume calculations, you ensure that your math is correct and the values you derive are accurate and usable.
This can make the difference between getting the right amount of chemicals for an experiment and making a potentially costly or dangerous mistake.