Question

What volume of \(0.0175 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\) must be added to \(50.0 \mathrm{mL}\) of \(0.0248 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\) so that the resulting solution has a molarity of exactly \(0.0200 \mathrm{M}\) ? Assume that the volumes are additive.

Short answer

To calculate \(V_2\), substitute the known values into the equation from Step 3: \(V_2 = ((0.0248 M * 50.0 mL) - (0.0200 M * 50.0 mL)) / (0.0200 M - 0.0175 M)\). After calculation, you find that \(V_2 \approx 136.36 mL\). It indicates that about 136.36 mL of \(0.0175 M CH_{3}OH\) must be added to the solution.

Step-by-step solution

Label known variables

Firstly, identify and label all the known variables in the problem. Here, \(M_1 = 0.0248 M\), \(V_1 = 50.0 mL\), \(M_2 = 0.0175 M\), and \(M_f = 0.0200 M\).

Calculate the final volume

Since the volumes are additive, the final volume of the solution will be the volume of the first solution plus the volume of the second solution. Here, the second volume \(V_2\) is not known. So, let's denote the final volume as \(V_f = V_1 + V_2\).

Solve for the missing variable

Use the dilution equation \(M_1 V_1 + M_2 V_2 = M_f V_f\) to solve for \(V_2\). Firstly, substitute \(V_f\) on the right-hand side with \(V_1 + V_2\) to get \(M_1 V_1 + M_2 V_2 = M_f (V_1 + V_2)\). Then, rearrange the equation to find \(V_2: V_2 = (M_1 V_1 - M_f V_1) / (M_f - M_2)\). Next, substitute the known variables into this equation to solve for \(V_2\).

Key concepts

Molarity

Molarity is an essential concept in chemistry, as it measures the concentration of a solution. Molarity, represented by the letter \(M\), tells us how many moles of a solute are present in one liter of solution. When we talk about a solution’s molarity, we essentially discuss how packed the solution is with solute particles.

In any solution preparation, knowing the molarity is crucial because it allows chemists and students to understand how solute particles interact in a mixture. It is calculated using the formula:
\[ M = \frac{n}{V} \]
where \(n\) is the number of moles of solute, and \(V\) is the volume of the solution in liters.

For example, if we have a solution of \(0.0248 \mathrm{M} \mathrm{CH}_3\mathrm{OH}\), it indicates that there are 0.0248 moles of \(\mathrm{CH}_3\mathrm{OH}\) in every liter of the solution. Understanding molarity enables us to prepare solutions to exact specifications, which is vital for maintaining consistency in experiments.

Solution Volume

Solution volume refers to the total space that a solution occupies. It is typically measured in liters or milliliters.

In our exercise, we need to determine the final volume of the overall solution to solve for the unknown variables. This is done by adding together the volumes of individual components involved in the dilution, under the assumption that volumes are additive.

Understanding solution volume is crucial when mixing solutions, as it directly affects the concentration and the resulting molarity. When volumes of different concentrations are combined, the total number of solute particles is dispersed over a new total volume. Thus, calculating the final volume accurately ensures consistent and reliable results, a necessity in any laboratory setting or quantitative analysis.

• Know your initial volumes, as they serve as the basis for additions.
• Remember, the sum of all constituent volumes equals the total or overall volume.

Dilution Equation

The dilution equation is an important tool for determining how to create a solution with a desired concentration from a more concentrated solution. This equation, \(M_1V_1 + M_2V_2 = M_fV_f\), connects the molarity and volume of the initial solutions to the molarity and volume of the final solution.

In our problem, this equation is pivotal for finding out how much of one solution we must add to another to achieve the target molarity of 0.0200 M. We start by identifying the known variables, which are \(M_1, V_1, M_2,\) and \(M_f\). Often, the missing piece will be a volume, which we solve for.

• First, substitute the known values into the dilution equation.
• Isolate the unknown variable, typically a volume \(V_2\).
• Solve for that variable to achieve the desired final concentration.

The dilution equation is straightforward, yet powerful, enabling accurate adjustments in solution concentration, critical in chemistry practices.