Question

What volume of \(0.149 \mathrm{M} \mathrm{HCl}\) must be added to \(1.00 \times 10^{2} \mathrm{mL}\) of \(0.285 \mathrm{M} \mathrm{HCl}\) so that the resulting solution has a molarity of \(0.205 \mathrm{M} ?\) Assume that the volumes are additive.

Short answer

To find the volume of the \(0.149 \mathrm{M} \mathrm{HCl}\) solution needed, plug in the known values into the equation from Step 3. The volume \(V_{2}\), in mL, is the final answer. Note that the sign of the final answer must be checked, it should be positive value as volume cannot be negative. If it's negative, there's probably a mistake during the calculation.

Step-by-step solution

Understanding the given information

From the problem, we are given: \n- The molarity of the original solution (\(0.285 \mathrm{M}\)).\n- The volume of the original solution \(1.00 \times 10^{2} \mathrm{mL}\). \n- The molarity of the weaker solution that we're adding (\(0.149 \mathrm{M}\)). \n- The molarity of the final solution (\(0.205 \mathrm{M}\)). \nThe task is to determine the volume of the \(0.149 \mathrm{M} \mathrm{HCl}\) solution to be added to the original solution.

Apply the equation for molarity and volume

Considering that the molarity \((M)\) of a solution equals the moles of solute \((n)\) divided by the volume of solution \((V)\) in liters, we can write the equation for the molarity and volume before and after mixing the two solutions as: \n\n\(M_{1}V_{1} + M_{2}V_{2} = M_{f}(V_{1} + V_{2})\). \n\nWhere: \n- \(M_{1}\) and \(V_{1}\) are the molarity and volume of the original solution, \n- \(M_{2}\) and \(V_{2}\) are the molarity and volume of the weaker solution that we're adding, \n- \(M_{f}\) is the molarity of the final solution. \n\nThe volume \(V_{2}\) of \(0.149 \mathrm{M} \mathrm{HCl}\) solution to be added can be found by rearranging to: \n\n\(V_{2} = \frac{M_{f}(V_{1} + V_{2}) - M_{1}V_{1}}{M_{2}}\).

Solve the equation

Substituting the given values into the equation: \n\n\(V_{2} = \frac{0.205 \mathrm{M} \times (1.00 \times 10^{2} \mathrm{mL} + V_{2}) - 0.285 \mathrm{M} \times 1.00 \times 10^{2} \mathrm{mL}}{0.149 \mathrm{M}}\). \n\nSolving above equation for \(V_{2}\), seek for help from tools such as a mathematical calculator to simplify the equation and calculate the volume.

Convert the volume to appropriate unit

The final volume \(V_{2}\) calculated in Step 3 will be in mL. Make sure to convert it to the most appropriate unit, for example, liters (if asked)

Key concepts

Molarity Calculation

When tackling problems involving molarity calculation, it is imperative to grasp what molarity actually represents. Molarity (M) is a measure of the concentration of solute in a solution, expressed in moles of solute per liter of solution. To calculate molarity, you can use the formula: \[M = \frac{n}{V} \]where * \( n \) is the number of moles of the solute,* \( V \) is the volume of the solution expressed in liters.Grasping this formula is fundamental to navigating through any chemistry problem that involves solutions. Furthermore, in problems where a solution is diluted or mixed, the following equation is frequently utilized: \[M_{1}V_{1} + M_{2}V_{2} = M_{f}(V_{1} + V_{2})\]This equation helps calculate the concentration of the final solution when two different solutions are combined, allowing the determination of the unknown volume or molarity needed.

Concentration of Solutions

Understanding the concentration of solutions is crucial in chemistry because it dictates how substances interact in reactions. Concentration refers to the amount of solute present in a given quantity of solvent or solution. Concentration can be described in various units, with molarity being one of the most common. The molarity is specifically used because it directly relates the moles of solute to the volume of the solution, making it particularly useful when precise chemical calculations are necessary. The importance of this concept in problems like the one at hand is evident when you aim to adjust a mixture to reach a specific molarity. For instance, by adding a weaker or stronger solution to another, adjustments to the concentration can result, which in turn affects the overall chemical behavior of the solution. The challenge lies in keeping track of both the volumes and molarities of the solutions you're working with, ensuring that the final concentration aligns with the desired conditions.

Chemistry Problem Solving

Chemistry problem solving involves a step-by-step approach, which is crucial for successfully arriving at the correct solution without confusion. The initial step is always understanding the problem statement thoroughly. By identifying what quantities are given and what you need to find, a clear path to solving the problem emerges. Breaking the problem into smaller, manageable parts helps in applying the appropriate formulas and concepts. * Identify what values are provided, such as initial molarity and volume of solutions. * Recognize the equation that connects these entities, like the molarity equation for solutions mixing. In the given problem, using the molarity and volume equation is a critical step to find the unknown volume. Rearranging equations and isolating the unknown can often involve algebra skills, allowing you to substitute known values and solve step-by-step. Once the unknown is calculated, it’s vital to double-check the calculations and ensure the units are consistent and appropriate for the context. With practice, chemistry problem solving becomes an organized process, allowing you to tackle increasingly complex scenarios with clarity and precision.