Question

A \(25-\mathrm{mL}\) sample of \(0.50 \mathrm{M} \mathrm{NaOH}\) is combined with a \(75-\mathrm{mL}\) sample of \(0.50 \mathrm{M} \mathrm{NaOH} .\) What is the concentration of the resulting \(\mathrm{NaOH}\) solution?

Short answer

The concentration of the resulting NaOH solution is 0.50 M.

Step-by-step solution

Understand the Problem

We are given two solutions of NaOH with different volumes but the same concentration, and we need to find the concentration of the resulting solution when they are combined.

Calculate Moles of NaOH in Each Solution

Use the formula \( ext{moles} = ext{volume (L)} \times ext{concentration (M)} \) to calculate the moles in each solution. For the 25 mL solution: \( 0.025 ext{ L} \times 0.50 ext{ M} = 0.0125 ext{ moles} \). For the 75 mL solution: \( 0.075 ext{ L} \times 0.50 ext{ M} = 0.0375 ext{ moles} \).

Calculate Total Moles of NaOH

Add the moles from both solutions together: \(0.0125 ext{ moles} + 0.0375 ext{ moles} = 0.0500 ext{ moles} \).

Calculate Total Volume of the Solution

Add the volumes of the two solutions: \(25 ext{ mL} + 75 ext{ mL} = 100 ext{ mL} \). Convert this to liters: \(100 ext{ mL} = 0.100 ext{ L} \).

Calculate the Concentration of the Resulting Solution

Use the formula \( ext{concentration (M)} = \frac{ ext{total moles}}{ ext{total volume (L)}} \) to find the concentration of the combined solution: \( \frac{0.0500 ext{ moles}}{0.100 ext{ L}} = 0.50 ext{ M} \).

Key concepts

Molarity Calculation

Calculating the molarity of a solution is a fundamental skill in chemistry. It involves knowing the number of moles of solute present in a specific volume of solvent, which is usually water in most chemical reactions. Molarity \((M)\) is defined as the number of moles of solute per liter of solution. The formula to calculate molarity is: \[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \] Understanding how to use this formula is crucial. It helps in determining concentrations in reactions, which impacts reaction rates, yields, and other properties. When working with solutions in the lab, always ensure your volume measurements are precise and your solute is completely dissolved for an accurate molarity reading. Breaking down this formula can make complex calculations seem more manageable.

Moles and Volume

To understand solutions better, we should grasp the relationship between moles and volume. The mole is a unit in chemistry used to express amounts of a chemical substance, similar to how "dozen" is used for quantities like eggs. When given the molarity and volume of a solution, you can find the number of moles using the formula: \[ \text{moles} = \text{volume (L)} \times \text{concentration (M)} \] For example, in the exercise, we find the moles of NaOH in each solution by multiplying their volume in liters by their concentration in molarity.

• **25 mL solution:** \(0.025 \, \text{L} \times 0.50 \, \text{M} = 0.0125 \, \text{moles}\)
• **75 mL solution:** \(0.075 \, \text{L} \times 0.50 \, \text{M} = 0.0375 \, \text{moles}\)

Having a good handle on this method allows precision in mixing and predicting product yields in laboratory settings.

Combining Solutions

When combining solutions, it is essential to calculate the resulting concentration accurately. This involves several steps that utilize your understanding of moles and volume. Start by finding the total moles of solute from each solution. You do this by adding up all the moles from the solutions you are combining. For instance, in this problem:\[ 0.0125 \, \text{moles} + 0.0375 \, \text{moles} = 0.0500 \, \text{moles} \]Next, determine the total volume of the combined solutions. Convert any volumes into liters if necessary before summing them:25 mL + 75 mL = 100 mL = 0.100 LWith the total moles and total volume known, compute the molarity of the final solution using the molarity formula:\[ \text{M} = \frac{0.0500 \, \text{moles}}{0.100 \, \text{L}} = 0.50 \, \text{M} \] Accurate combination and calculation ensure that solutions behave as expected in chemical reactions, highlighting the significance of precision in chemical processes.