Question

Calculate each of the following quantities: (a) Volume (L) of \(2.050 \mathrm{M}\) copper(II) nitrate that must be diluted with water to prepare \(750.0 \mathrm{~mL}\) of a \(0.8543 \mathrm{M}\) solution (b) Volume (L) of \(1.63 M\) calcium chloride that must be diluted with water to prepare \(350 . \mathrm{mL}\) of a \(2.86 \times 10^{-2} \mathrm{M}\) chloride ion solution (c) Final volume (L) of a \(0.0700 \mathrm{M}\) solution prepared by diluting \(18.0 \mathrm{~mL}\) of \(0.155 \mathrm{M}\) lithium carbonate with water

Short answer

a) 0.3127 L, b) 0.00307 L, c) 0.0399 L

Step-by-step solution

Understand the relationship between concentrations and volumes

For dilution problems, use the dilution formula: \[ M_1 V_1 = M_2 V_2 \] where \( M_1 \) and \( V_1 \) are the initial concentration and volume, and \( M_2 \) and \( V_2 \) are the final concentration and volume.

Solve for the initial volume for part (a)

Given: \( M_1 = 2.050 \text{ M} \), \( M_2 = 0.8543 \text{ M} \), and \( V_2 = 750.0 \text{ mL} \). Rearrange the dilution formula to solve for \( V_1 \): \[ V_1 = \frac{M_2 V_2}{M_1} \] Substitute the known values: \[ V_1 = \frac{0.8543 \text{ M} \times 750.0 \text{ mL}}{2.050 \text{ M}} = 312.7 \text{ mL} \]

Convert volume to liters for part (a)

Convert 312.7 mL to liters by dividing by 1000: \[ V_1 = 312.7 \text{ mL} \div 1000 = 0.3127 \text{ L} \]

Solve for the initial volume for part (b)

First, find the concentration of calcium chloride required to give the given chloride ion concentration. For \( \text{CaCl}_2 \), each unit provides 2 chloride ions. Therefore, \[ M_{CaCl_2} = \frac{2.86 \times 10^{-2} \text{ M}}{2} = 1.43 \times 10^{-2} \text{ M} \] Use the dilution formula: Given \( M_1 = 1.63 \text{ M} \), \( M_2 = 1.43 \times 10^{-2} \text{ M} \), and \( V_2 = 350 \text{ mL} \): \[ V_1 = \frac{M_2 V_2}{M_1} = \frac{1.43 \times 10^{-2} \text{ M} \times 350 \text{ mL}}{1.63 \text{ M}} = 3.07 \text{ mL} \]

Convert volume to liters for part (b)

Convert 3.07 mL to liters by dividing by 1000: \[ V_1 = 3.07 \text{ mL} \div 1000 = 0.00307 \text{ L} \]

Solve for the final volume for part (c)

Given \( M_1 = 0.155 \text{ M} \), \( V_1 = 18.0 \text{ mL} \), and \( M_2 = 0.0700 \text{ M} \): Rearrange the formula to solve for \( V_2 \): \[ V_2 = \frac{M_1 V_1}{M_2} \] Substitute the known values: \[ V_2 = \frac{0.155 \text{ M} \times 18.0 \text{ mL}}{0.0700 \text{ M}} = 39.9 \text{ mL} \]

Convert volume to liters for part (c)

Convert 39.9 mL to liters by dividing by 1000: \[ V_2 = 39.9 \text{ mL} \div 1000 = 0.0399 \text{ L} \]

Key concepts

molarity

Molarity is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution (mol/L). It's represented by the symbol M. Understanding molarity helps you to know how much of a substance is present in a given volume of liquid. For example, if you have a 1 M solution of sodium chloride (NaCl), it means there is 1 mole of NaCl dissolved in 1 liter of the solution.
To calculate molarity, use the formula:
\[ M = \frac{n}{V} \]
where:

\t
• M is molarity
\t
• n is the number of moles of solute
\t
• V is the volume of the solution in liters

Knowing the molarity of solutions is crucial in many chemical calculations, including stoichiometry and dilutions.

solution concentration

Solution concentration refers to the amount of solute that is dissolved in a given volume of solvent. Concentrations can be expressed in various ways, including molarity (M), parts per million (ppm), and percent composition by mass or volume. In dilution calculations, molarity is the most commonly used unit.
When dealing with different solutions, it's important to compare their concentrations to understand their relative strengths. For example, a 3 M solution of hydrochloric acid (HCl) is more concentrated than a 1 M solution of HCl.
Understanding solution concentration is vital in preparing solutions for experiments, ensuring the proper reaction conditions, and achieving the desired results in both laboratory and industrial settings.

volume conversion

Volume conversion is necessary when working with solutions and dilutions, especially when volumes are given in different units. The most common volume units in chemistry are liters (L) and milliliters (mL), and knowing how to convert between them is essential. To convert milliliters to liters, divide by 1000, since there are 1000 milliliters in one liter.
For example, converting 312.7 mL to liters involves:
\[ 312.7 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.3127 \text{ L} \]
Volume conversions are also essential when preparing solutions and performing calculations that require consistent units throughout the problem.

dilution formula

The dilution formula is a key equation used when you need to dilute a concentrated solution to a lower concentration. The formula is:
\[ M_1 V_1 = M_2 V_2 \]
where:

\t
• M1 is the initial molarity
\t
• V1 is the initial volume
\t
• M2 is the final molarity
\t
• V2 is the final volume

This equation shows that the product of the initial concentration and volume (before dilution) is equal to the product of the final concentration and volume (after dilution). For instance, if you have 2.050 M copper(II) nitrate and want to prepare 750.0 mL of 0.8543 M solution, you would rearrange the formula to solve for V1 (the initial volume needed):
\[ V_1 = \frac{M_2 V_2}{M_1} \]
Substituting the given values:
\[ V_1 = \frac{0.8543 \text{ M} \times 750.0 \text{ mL}}{2.050 \text{ M}} = 312.7 \text{ mL} \]
The dilution formula is indispensable in laboratories and many scientific applications where specific solution concentrations are needed.