Problem 74 - Chemistry The Central Science In Si Units

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) 9.17 mL; (b) 0.240 M.

Step 1: Identify the Given Data for Part (a)

From the problem we know: The stock solution concentration is \(6.0 \, \text{M HNO}_3\), and the desired concentration for the dilution is \(0.500 \, \text{M HNO}_3\). We need to prepare \(110 \, \text{mL}\) of this diluted solution.

Step 2: Determine the Relationship to Use

The dilution equation is \(C_1V_1 = C_2V_2\), where \(C_1\) and \(V_1\) are the concentration and volume of the stock solution, and \(C_2\) and \(V_2\) are the concentration and volume after dilution.

Step 3: Calculate the Volume Needed for Part (a)

Rearrange the dilution equation to solve for \(V_1\): \[ V_1 = \frac{C_2V_2}{C_1} = \frac{0.500 \, \text{M} \times 110 \, \text{mL}}{6.0 \, \text{M}} \]Calculate: \[ V_1 = \frac{55 \, \text{mL M}}{6.0 \, \text{M}} = 9.17 \, \text{mL} \]

Step 4: Identify the Given Data for Part (b)

You start with \(10.0 \, \text{mL}\) of a \(6.0 \, \text{M HNO}_3\) stock solution and dilute it to a final volume of \(0.250 \, \text{L}\).

Step 5: Determine the New Concentration After Dilution

Use the dilution formula again, \(C_1V_1 = C_2V_2\). Substitute known values to solve for \(C_2\): \[ C_2 = \frac{C_1V_1}{V_2} = \frac{6.0 \, \text{M} \times 10.0 \, \text{mL}}{250 \, \text{mL}} \]Calculate: \[ C_2 = \frac{60 \, \text{mL M}}{250 \, \text{mL}} = 0.240 \, \text{M} \]

Key Concepts

Concentration Calculation

Understanding the concept of concentration is key to performing solution-related calculations. In chemistry, concentration typically refers to the amount of solute dissolved in a unit volume of solution. It helps us understand how strong or weak a solution is.
For practical purposes, concentrations are often given in molarity (M), where 1 M means one mole of solute per liter of solution. Calculating concentration requires a clear understanding of this relation between moles, volume, and molarity:

• Moles of solute: How much of the chemical is actually in the solution.
• Volume of the solution: The space it occupies, usually measured in liters.
• Molarity: A measure of concentration itself, usually expressed in moles per liter (mol/L).

By rearranging the formula, we ensure precise measurement of how much stock solution we need to dilute or how concentrated a new solution will be after dilution.

Dilution Equation

The dilution equation is an essential tool in chemistry, especially when working with solutions. It relates the concentration and volume of a stock solution to the concentration and volume after dilution. The equation is:
\[C_1V_1 = C_2V_2\] Here:

• \(C_1\) is the initial concentration of the stock solution.
• \(V_1\) is the volume of the stock solution used.
• \(C_2\) is the concentration after dilution.
• \(V_2\) is the total volume of the solution after dilution.

Using this equation, we can find any one of the variables if the other three are known. In practice, it helps us adjust how concentrated we want the solution to be by directly changing the amount of solvent or stock solution used, ensuring flexibility in laboratory settings.

Molarity

Molarity, a standard unit of concentration, indicates how many moles of solute are dissolved in one liter of solution. It helps chemists understand the precise amount of a substance within a given solution.
Molarity is expressed as \(M\), which equals moles of solute per liter of solution:\[M = \frac{\text{moles of solute}}{\text{liters of solution}} \]When preparing solutions or conducting titrations, knowing the molarity allows accurate and reproducible results. However, when diluting, it is crucial to remember that molarity will change as the volume changes due to added solvent, while the number of moles of solute remains constant.
Molarity simplifies calculations and communications in chemistry, making it a go-to metric for expressing concentration in both educational contexts and professional laboratories.