Question

Calculate the new molarity when \(150 . \mathrm{mL}\) of water is added to each of the following solutions. a. \(125 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HBr}\) b. \(155 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}\) c. \(0.500 \mathrm{~L}\) of \(0.250 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) d. \(15 \mathrm{~mL}\) of \(18.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\)

Short answer

The new molarities for the given solutions are: a) \(0.091 M\), b) \(0.127 M\), c) \(0.192 M\), and d) \(1.64 M\).

Step-by-step solution

Problem a.

First, calculate the final volume of the solution: \(V_2 = 125 mL + 150 mL = 275 mL\) Next, use the dilution formula to calculate the new molarity: \(M_1V_1=M_2V_2\) \(0.200 M \times 125 mL = M_2 \times 275 mL\) Divide both sides by 275 mL: \(M_2 = \frac{0.200 M \times 125 mL}{275 mL}= 0.091 M\) Answer: The new molarity for solution a is \(0.091 M\).

Problem b.

First, calculate the final volume of the solution: \(V_2 = 155 mL + 150 mL = 305 mL\) Next, use the dilution formula to calculate the new molarity: \(M_1V_1=M_2V_2\) \(0.250 M \times 155 mL = M_2 \times 305 mL\) Divide both sides by 305 mL: \(M_2 = \frac{0.250 M \times 155 mL}{305 mL}= 0.127 M\) Answer: The new molarity for solution b is \(0.127 M\).

Problem c.

First, convert the initial volume from liters to milliliters: \(0.500 L = 500 mL\) Next, calculate the final volume of the solution: \(V_2 = 500 mL + 150 mL = 650 mL\) Now, use the dilution formula to calculate the new molarity: \(M_1V_1=M_2V_2\) \(0.250 M \times 500 mL = M_2 \times 650 mL\) Divide both sides by 650 mL: \(M_2 = \frac{0.250 M \times 500 mL}{650 mL}= 0.192 M\) Answer: The new molarity for solution c is \(0.192 M\).

Problem d.

First, calculate the final volume of the solution: \(V_2 = 15 mL + 150 mL = 165 mL\) Next, use the dilution formula to calculate the new molarity: \(M_1V_1=M_2V_2\) \(18.0 M \times 15 mL = M_2 \times 165 mL\) Divide both sides by 165 mL: \(M_2 = \frac{18.0 M \times 15 mL}{165 mL}= 1.64 M\) Answer: The new molarity for solution d is \(1.64 M\).

Key concepts

Molarity Calculations

In chemistry, molarity is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution. Calculating molarity is essential for solution preparation and chemical reactions. Molarity, denoted by the symbol \( M \), combines the amount of substance (in moles) with the volume of the solution. The formula for calculating molarity is:

• \( M = \frac{n}{V} \)

Where \( n \) is the number of moles of solute, and \( V \) is the volume of the solution in liters.

In exercises like the one provided, understanding molarity helps us determine how diluted or concentrated a solution becomes after adding more solvent, such as water. This concept becomes particularly relevant when performing experiments that require specific concentrations.

Knowing how to calculate molarity allows chemists and students alike to prepare solutions accurately and ensure that reactions proceed as expected.

Dilution Formula

Dilution is the process of reducing the concentration of a solute in a solution, usually by adding more solvent. The formula for dilution is based on the principle that the amount of solute remains constant before and after dilution, represented by:

• \( M_1V_1 = M_2V_2 \)

Where:

• \( M_1 \) is the initial molarity
• \( V_1 \) is the initial volume
• \( M_2 \) is the final molarity
• \( V_2 \) is the final volume

The dilution formula allows us to calculate one of these unknowns if the other three are provided.

In the context of the original exercise, understanding dilution is crucial for accurately determining how the addition of \( 150 \) mL of water changes the molarity of several solutions.

For each problem, the final volume \( V_2 \) is the sum of the initial volume \( V_1 \) and the volume of added water. By rearranging the dilution formula and solving for the new molarity \( M_2 \), students can find how concentrated or diluted the solution becomes.

Chemistry Problem Solving

Solving chemistry problems effectively requires a clear understanding and correct application of principles and formulas. In tasks such as determining the new molarity after a dilution, systematic problem-solving approaches are instrumental.Here are steps often followed:

• Identify and list known quantities: initial molarity \( M_1 \), initial volume \( V_1 \), and added volume of solvent.
• Calculate the total final volume \( V_2 \) by adding \( V_1 \) and the added solvent volume.
• Apply the dilution formula \( M_1V_1 = M_2V_2 \) to solve for the unknown new molarity \( M_2 \).
• Allewa perform checks: Revisiting each step can help catch potential errors and yield confidence in the solution.

Conceptual clarity is important. It helps not only in solving single problems but also in grasping how to approach unfamiliar questions.

With practice, chemistry problem-solving becomes a skill, allowing students to tackle varied issues ranging from academic exercises to real-world applications in laboratories.