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 \(\mathrm{M} \mathrm{HNO}_{3} ?\) (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?

Short answer

(a) To prepare 110 mL of 0.500 M HNO3 solution, we need 9.17 mL of the 6.0 M HNO3 stock solution. (b) The concentration of the diluted solution when 10.0 mL of the stock solution is diluted to 0.250 L is 0.240 M.

Step-by-step solution

Part (a): Finding the volume of the stock solution

Use the dilution formula to find the volume of the stock solution required to prepare the 0.500 M HNO3 solution: M1V1 = M2V2 We know: M1 = 6.0 M (concentration of the stock solution) M2 = 0.500 M (concentration of the diluted solution) V2 = 110 mL (volume of the diluted solution) Rearrange the formula to find V1: V1 = (M2V2) / M1 Plug in the given values: V1 = (0.500 M × 110 mL) / 6.0 M Calculate the value of V1: V1 = 9.17 mL So, to prepare 110 mL of 0.500 M HNO3 solution, we need 9.17 mL of the 6.0 M HNO3 stock solution.

Part (b): Finding the concentration of the diluted solution

Use the dilution formula to find the concentration of the diluted solution when 10.0 mL of the stock solution is diluted to 0.250 L: M1V1 = M2V2 We know: M1 = 6.0 M (concentration of the stock solution) V1 = 10.0 mL (volume of the stock solution) V2 = 0.250 L = 250 mL (volume of the diluted solution) Rearrange the formula to find M2: M2 = (M1V1) / V2 Plug in the given values: M2 = (6.0 M × 10.0 mL) / 250 mL Calculate the value of M2: M2 = 0.240 M So, the concentration of the diluted solution when 10.0 mL of the stock solution is diluted to 0.250 L is 0.240 M.

Key concepts

Concentration

Concentration refers to how much of a substance is present in a certain volume of a solution. It is often expressed in terms of molarity, but there are other ways to define it too, like percentage or mass per volume.
Understanding concentration is crucial because it tells us how strong or weak a solution is.
In chemistry, knowing how to calculate and manipulate concentration can help you achieve the desired effects in reactions and applications.

• Concentration can dictate the rate of chemical reactions. Higher concertation can mean faster reactions.
• It is important in safety, as some substances can be harmful or volatile at higher concentrations.

For instance, in the exercise, we use the concept of concentration to determine how much of a stock solution is needed to create a new solution with a specific concentration. The stock solution is more concentrated, meaning it has more solute particles in a smaller volume.

Molarity

Molarity is one way to express concentration, specifically as moles of solute per liter of solution. This is a common unit of concentration because it directly relates the number of molecules or atoms in the solution to its volume.

• Molarity is expressed in units of molarity (M), which is moles per liter.

To calculate molarity, you divide the number of moles of solute by the volume of the solution in liters. This is essential in chemistry as it provides a direct relationship between volume and the number of solute particles.
In our original problem, we see molarity in action:

• The stock solution has a molarity of 6.0 M.
• The desired diluted solution has a molarity of 0.500 M or 0.240 M depending on the scenario.

Through these calculations, we can understand how much of the stock solution is needed to achieve our target molarity after dilution.

Solution Preparation

Preparing a solution involves mixing solutes with solvents to yield a homogenous liquid mixture.
The goal is to achieve a desired concentration for use in experiments or industrial processes.
Steps to prepare a solution:
1. **Calculate the required solute's amount**: Use formulas or given concentrations to determine how much solute you need to add to achieve the desired final concentration.
2. **Measure and mix**: Accurately measure both the solute and the solvent. Combine them and allow the solute to dissolve completely.
3. **Adjust the volume if needed**: Ensure the final volume matches the desired amount, sometimes requiring a final addition of the solvent.
In our example, using the dilution formula \( M_1V_1 = M_2V_2 \) helps calculate exactly how much stock solution to use or to predict the new concentration when diluting. This helps ensure consistency and accuracy in experiments, which is essential for replicable scientific results.