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 \(10.0\)-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?

Short answer

(a) To make a 1000.0 mL solution of 0.250 M NH3, you should dilute 16.9 mL of the 14.8 M NH3 stock solution. (b) The concentration of the final solution when you take a 10.0 mL portion of the stock solution and dilute it to a total volume of 0.500 L will be 0.296 M NH3.

Step-by-step solution

Set up the dilution formula for part (a)

The dilution formula is M1V1 = M2V2, where M1 = 14.8 M, V1 = volume of stock solution (in mL), M2 = 0.250 M, and V2 = 1000.0 mL.

Solve for V1 for part (a)

Plug in the given values into the dilution formula and solve for V1: \(14.8\, \mathrm{M} \times V1 = 0.250\, \mathrm{M} \times 1000.0\, \mathrm{mL}\) Divide both sides by 14.8 M: \(V1 = \frac{0.250\, \mathrm{M} \times 1000.0\, \mathrm{mL}}{14.8\, \mathrm{M}}\) Calculate the value for V1: \(V1 = 16.9\, \mathrm{mL}\) Answer (a): To make a 1000.0 mL solution of 0.250 M NH3, you should dilute 16.9 mL of the 14.8 M NH3 stock solution. (b)

Set up the dilution formula for part (b)

The dilution formula is M1V1 = M2V2, where M1 = 14.8 M, V1 = 10.0 mL, M2 is the unknown concentration of the final solution, and V2 = 0.500 L or 500 mL.

Solve for M2 for part (b)

Plug in the given values into the dilution formula and solve for M2: \(14.8\, \mathrm{M} \times 10.0\, \mathrm{mL} = M2 \times 500\, \mathrm{mL}\) Divide both sides by 500 mL: \(M2 = \frac{14.8\, \mathrm{M} \times 10.0\, \mathrm{mL}}{500\, \mathrm{mL}}\) Calculate the value for M2: \(M2 = 0.296\, \mathrm{M}\) Answer (b): The concentration of the final solution when you take a 10.0 mL portion of the stock solution and dilute it to a total volume of 0.500 L will be 0.296 M NH3.

Key concepts

Dilution Formula

Understanding how to dilute a concentrated solution to a specified molarity is an essential laboratory skill. The dilution formula, which is instrumental for such calculations, is given by \( M1V1 = M2V2 \). Here, \( M1 \) and \( V1 \) represent the molarity and volume of the initial concentrated solution, while \( M2 \) and \( V2 \) are the molarity and volume of the final diluted solution, respectively.

This simple yet powerful equation allows you to calculate the volume of the original concentrated solution (stock solution) needed to achieve a desired concentration after dilution. To visualize this, imagine you have a very strong coffee (high molarity) and you wish to make it milder (lower molarity) by adding some water (increasing volume). The concept is the same in chemistry when dealing with solutions.

Molarity

Molarity, abbreviated as \( M \), is a unit used to describe the concentration of a solution. It is defined as the number of moles of a solute per liter of solution. In other words, \( 1 M \) solution contains one mole of a solute dissolved in one liter of solution (which includes both solute and solvent). Understanding molarity is crucial when preparing solutions in chemistry for reactions or dilutions because it helps you to ensure that the reaction mixture has the correct amount of reactants.

When it comes to dilutions, molarity helps you determine how much you’re weakening your solution. If you have a high molarity, you have a concentrated solution; if you have a low molarity, you have a dilute solution. Molarity can also guide you in stoichiometric calculations, where reactants must be in specific proportions.

Volume Conversion

Volume conversion is a necessity in laboratory work, especially when measurements need to be converted from one unit to another. It is common for some procedures to require milliliters (mL), while others may call for liters (L). Remember, \( 1000 mL = 1 L \) — this simple ratio is key in converting volumes. It's crucial when performing dilution calculations because the dilution formula requires consistency in units.

For example, if a solution needs to be prepared with a certain volume in liters, but your stock solution is measured in milliliters, you have to convert these measurements to match. Knowing how to swiftly convert between these units without errors is essential to ensure the accuracy and consistency of the results in any chemical procedure.

Concentration of Solutions

Concentration of solutions describes how much solute is present in a given quantity of solvent or solution. There are various ways to express this, with molarity being just one of them. Working with concentration is like following a recipe - too much or too little of an ingredient can change the outcome. In chemistry, too much or too little of a reactant can affect the rate of reaction and the yield of the product.

In dilution problems, understanding the initial concentration of a stock solution helps you determine how much you can ‘water down’ a solution while achieving the desired concentration. For instance, in our examples, the stock solution of \( NH_3 \) has a high concentration, but we want to achieve a solution of lower concentration. The process of reaching that final concentration involves carefully measuring volumes based on the concentrations, which is where the dilution formula is applied.