Question

(a) How would you prepare \(175.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{AgNO}_{3}\) solution starting with pure \(\mathrm{AgNO}_{3} ?\) (b) An experiment calls for you to use \(100 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HNO}_{3}\) solution. All you have available is a bottle of \(3.6 \mathrm{M} \mathrm{HNO}_{3} .\) How would you prepare the desired solution?

Short answer

(a) To prepare \(175.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{AgNO}_{3}\) solution, weigh 4.456 grams of pure AgNO3 and dissolve it in distilled water. Gradually add more distilled water, mixing the solution until you reach a final volume of 175.0 mL. (b) To prepare \(100 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HNO}_{3}\) solution from a \(3.6 \mathrm{M} \mathrm{HNO}_{3}\) solution, measure 13.89 mL of the 3.6 M HNO3 solution and pour it into a suitable container. Gradually add distilled water, mixing the solution until you reach a final volume of 100 mL.

Step-by-step solution

Calculate the moles of solute needed

We need to find the moles of AgNO3 required to prepare the solution. We will use the equation: moles = volume x concentration moles of AgNO3 = \(175.0 \mathrm{~mL} \times 0.150 \mathrm{M}\) moles of AgNO3 = \(26.25 \mathrm{~mmol}\)

Convert moles to grams

Now, we will convert the moles to grams using the molar mass of AgNO3. Molar mass of AgNO3 = \(169.87 \mathrm{~g/mol}\) grams of AgNO3 = moles × molar mass grams of AgNO3 = \(26.25 \mathrm{~mmol} \times \frac{169.87 \mathrm{~g}}{1000 \mathrm{~mmol}}\) grams of AgNO3 = \(4.456 \mathrm{~g}\)

Prepare the solution

Weigh 4.456 grams of pure AgNO3 and dissolve it in distilled water. Gradually add more distilled water, mixing the solution until you reach a final volume of 175.0 mL. (b) Prepare \(100 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HNO}_{3}\) solution from a \(3.6 \mathrm{M} \mathrm{HNO}_{3}\) solution.

Apply the dilution equation

We will use the equation \(C_1V_1 = C_2V_2\), where \(C_1\) is the initial concentration, \(V_1\) is the initial volume, \(C_2\) is the final concentration, and \(V_2\) is the final volume. \(3.6 \mathrm{M} \times V_1 = 0.50 \mathrm{M} \times 100 \mathrm{~mL}\)

Calculate the initial volume

Now, we will solve for the initial volume, \(V_1\). \(V_1 = \frac{0.50 \mathrm{M} \times 100 \mathrm{~mL}}{3.6 \mathrm{M}}\) \(V_1 = 13.89 \mathrm{~mL}\) (approximately)

Prepare the solution

Measure 13.89 mL of the 3.6 M HNO3 solution and pour it into a suitable container. Gradually add distilled water, mixing the solution until you reach a final volume of 100 mL.

Key concepts

Molarity Calculation

Molarity is a term used in chemistry to describe the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. The formula to calculate molarity (M) is:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Understanding molarity is crucial for preparing solutions of desired concentrations. For example, to make a 0.150 M AgNO3 solution, the first step is to calculate the moles of AgNO3 needed. Since molarity is moles per liter, we must ensure that the volume is expressed in liters:
\[ \text{volume in liters} = 175.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} \]
Once the volume is in the correct unit, multiply by the molarity to find the moles of AgNO3:
\[ \text{moles of AgNO3} = \text{volume in liters} \times 0.150 \, M \]
This molarity calculation allows you to quantify the exact amount of solute needed to prepare a solution with a specific concentration.

Dilution Equation

The dilution equation is a powerful tool in solution preparation chemistry, allowing chemists to calculate the volume of a concentrated stock solution (also known as the initial solution) needed to achieve a desired concentration after dilution. The central formula is:
\[ C_1V_1 = C_2V_2 \]
where:\[ \begin{align*}C_1 & = \text{initial concentration (M)} \ V_1 & = \text{initial volume (L or mL)} \ C_2 & = \text{final concentration after dilution (M)} \ V_2 & = \text{final volume after dilution (L or mL)}\end{align*} \]
To use this equation, three of the four values need to be known. For instance, to dilute a 3.6 M HNO3 solution to a 0.50 M solution, we need to first state the final desired volume (100 mL). By rearranging the dilution equation to solve for the unknown initial volume (V1), we can determine exactly how much of the concentrated solution we should start with. After finding this volume, simply add enough solvent (usually water) to reach the total final volume required. This is an invaluable process for creating solutions of lesser intensity without the need to measure out specific masses of solute.

Molar Mass Conversion

Molar mass conversion is the process of converting between the amount of substance in moles and its mass in grams. It is an essential skill in chemistry as it allows for the preparation of a solution with a precise concentration. Molar mass is the mass of one mole of a substance (solute) and is expressed in grams per mole (g/mol).
To convert moles to grams, the formula is simple:
\[ \text{grams} = \text{moles} \times \text{molar mass} \]
Using AgNO3 as an example, if the moles of AgNO3 needed have been calculated to be 26.25 mmol (where 1 mmol is 0.001 mol), and the molar mass of AgNO3 is 169.87 g/mol, the mass in grams required can be found by:
\[ \text{grams of AgNO3} = 26.25 \, \text{mmol} \times \frac{169.87 \, \text{g/mol}}{1000 \, \text{mmol/mol}} \]
This results in the need for 4.456 g of AgNO3 to reach the desired concentration in the final solution volume. Properly understanding molar mass conversion is integral for chemists to weigh out the precise amount of substance needed to create a solution of a specific molarity.