Question

A \(15.00-\mathrm{mL}\) solution of potassium nitrate \(\left(\mathrm{KNO}_{3}\right)\) was diluted to \(125.0 \mathrm{~mL},\) and \(25.00 \mathrm{~mL}\) of this solution was then diluted to \(1.000 \times 10^{3} \mathrm{~mL}\). The concentration of the final solution is \(0.00383 \mathrm{M}\). Calculate the concentration of the original solution.

Short answer

The original solution concentration is 1.277 M.

Step-by-step solution

Determine dilutions

We begin by breaking down the dilution process into two main steps. Firstly, the solution is diluted from 15.00 mL to 125.0 mL, and secondly, from 25.00 mL of this new solution to 1000.0 mL.

Calculate intermediate concentration

To find the intermediate concentration after the first dilution, use the dilution formula: \[ C_1 V_1 = C_2 V_2 \]where \(C_1\) is the original concentration, \(V_1 = 15.0 \ \mathrm{mL}\), \(V_2 = 125.0 \ \mathrm{mL}\), and \(C_2\) is the concentration after dilution.Rearranging for \(C_2\), we get:\[ C_2 = \frac{C_1 \times V_1}{V_2} \]

Calculate final concentration after second dilution

After determining \(C_2\), which is the concentration after the first dilution, perform the second dilution using the equation:\[ C_2 V_3 = C_3 V_4 \]where \(V_3 = 25.0 \ \mathrm{mL}\), \(V_4 = 1000.0 \ \mathrm{mL}\), and \(C_3 = 0.00383 \ \mathrm{M}\).Rearrange for \(C_2\): \[ C_2 = \frac{C_3 \times V_4}{V_3} \]

Solve for original concentration

First, solve for \(C_2\) using the values given:\(C_3 = 0.00383 \ \mathrm{M}\), \(V_3 = 25.0 \ \mathrm{mL}\), and \(V_4 = 1000.0 \ \mathrm{mL}\).\[ C_2 = \frac{0.00383 \times 1000}{25} = 0.1532 \ \mathrm{M}\]Then, plug this \(C_2\) value back into the first dilution equation:\[ C_1 = \frac{C_2 \times V_2}{V_1} \] where \(V_2 = 125.0 \ \mathrm{mL}\) and \(V_1 = 15.0 \ \mathrm{mL}\).\[ C_1 = \frac{0.1532 \times 125}{15} = 1.277 \ \mathrm{M}\]

Concentration of original solution

The concentration of the original potassium nitrate solution is \(1.277 \ \mathrm{M}\).

Key concepts

Concentration Calculation

Understanding how to calculate concentration is fundamental in chemistry. Concentration is a measure of the amount of solute in a given volume of solution. It helps us understand how strong or weak a solution is in terms of solute presence. For example, when talking about a potassium nitrate solution, concentration indicates how much potassium nitrate is present in a solution.To determine concentration, you typically use the formula:\[C = \frac{n}{V}\]where \(C\) is concentration, \(n\) is the number of moles of the solute, and \(V\) is the volume of the solution in liters. In the given exercise, you need to calculate concentrations at different stages of dilution using slightly modified equations. This involves the use of known values of volume and concentration to determine unknowns. Each dilution step changes how much of the solute remains in a particular volume of solvent.

Dilution Formula

The dilution formula is a powerful tool in chemistry to calculate the concentration of a solution after it has been diluted. It follows the idea of conservation of mass, meaning the amount of solute remains constant before and after dilution. The formula is expressed as:\[C_1 V_1 = C_2 V_2\]Here, \(C_1\) and \(V_1\) are the concentration and volume of the concentrated solution, respectively, while \(C_2\) and \(V_2\) represent the concentration and volume after dilution. It’s crucial to remember that while the concentration may change, the total amount of solute does not. In the exercise, this formula is used twice: first, to find the intermediate concentration \(C_2\) after the initial dilution, and second, to determine \(C_2\) before it is used to solve for the final concentration \(C_3\). This step-by-step breakdown ensures the accurate calculation of the solution's strength at each stage.

Potassium Nitrate Solution

Potassium nitrate, a salt commonly used in fertilizers and food preservation, is represented by the formula \(KNO_3\). When creating a solution of potassium nitrate, it dissolves in water to release potassium (\(K^+\)) and nitrate (\(NO_3^-\)) ions. These ions' concentrations affect the solution's overall molarity.In this problem, a potassium nitrate solution undergoes sequential dilutions. As you work with potassium nitrate solutions, understanding the dilution's impact on the concentration of its ions is critical. In each dilution, the number of potassium and nitrate ions remains constant, even as the solution is more spread out due to increased volume. This point reinforces the importance of using precise volumes and concentrations throughout the calculation.

Molarity

Molarity is a way to express the concentration of a solution. It describes the number of moles of a solute per liter of solution, often denoted as \(M\). This unit of measurement allows chemists to easily convey how concentrated a solution is.Formulaically, molarity is expressed as:\[M = \frac{n}{V}\]where \(n\) is the number of moles of solute and \(V\) is the volume of the solution in liters. Using molarity helps to calculate how much of a solute is needed to achieve a desired concentration, or conversely, it can help find the concentration of a solution after changes in volume.In the provided example, the final molarity after the series of dilutions is given as \(0.00383 \, M\). To solve backwards to the initial concentration, you use this final molarity and apply the dilution formula, thereby ensuring each calculation uses consistent units for accuracy.