Quantitative Chemistry: Concentration and Solubility

Concentration of Solutions

The concentration of a solution is a measure of how many solute particles are present in a specific volume of solution. In quantitative chemistry, concentration is most commonly expressed in one of two ways: the amount in moles of a solute present in 1 $\text{dm}^3$ of solution ($\text{mol dm}^{-3}$) or the mass of solute in 1 $\text{dm}^3$ of solution ($\text{g dm}^{-3}$).

It is important to note that these measurements are based on the volume of the final solution, not the volume of the solvent. This is because the total volume often changes slightly when a solute dissolves. When performing calculations, volumes are frequently provided in $\text{cm}^3$. Since there are 1000 $\text{cm}^3$ in 1 $\text{dm}^3$, you must divide the volume in $\text{cm}^3$ by 1000 to convert it to $\text{dm}^3$.

Calculating Concentration from Mass

To calculate the concentration in $\text{g dm}^{-3}$, use the following relationship:

$\text{Concentration (g dm}^{-3}\text{)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (dm}^3\text{)}}$

Worked Example: Exercise 37

What is the concentration, in $\text{g dm}^{-3}$, of a solution containing 8.0 g of sodium hydroxide dissolved in 5.0 $\text{dm}^3$ of solution?

$\text{Concentration} = \frac{8.0\text{ g}}{5.0\text{ dm}^3} = 1.6\text{ g dm}^{-3}$

Worked Example: Exercise 38

Normal saline has a concentration of 9.0 $\text{g dm}^{-3}$ of $\text{NaCl}$. Calculate the mass of sodium chloride required to make 250 $\text{cm}^3$ of this solution.

1. Convert volume to $\text{dm}^3$: $\frac{250}{1000} = 0.250\text{ dm}^3$.

2. Calculate mass: $9.0\text{ g dm}^{-3} \times 0.250\text{ dm}^3 = 2.25\text{ g}$.

Calculating Molar Concentration

To find the concentration in $\text{mol dm}^{-3}$ (often called molarity), you first determine the number of moles of solute and then divide by the volume.

Worked Example: Exercise 40

What is the concentration, in $\text{mol dm}^{-3}$, of a solution formed when 5.6 g of potassium hydroxide ($\text{KOH}$) is dissolved in 500 $\text{cm}^3$?

1. Find the molar mass ($M_r$) of $\text{KOH}$: $39 + 16 + 1 = 56\text{ g mol}^{-1}$.

2. Calculate the number of moles: $\frac{5.6\text{ g}}{56\text{ g mol}^{-1}} = 0.10\text{ mol}$.

3. Convert volume: $\frac{500}{1000} = 0.500\text{ dm}^3$.

4. Calculate concentration: $\frac{0.10\text{ mol}}{0.500\text{ dm}^3} = 0.20\text{ mol dm}^{-3}$.

Rearranging the Concentration Formula

If you know the concentration and the volume, you can find the amount of solute:

$\text{Amount of solute (mol)} = \text{volume (dm}^3\text{)} \times \text{concentration (mol dm}^{-3}\text{)}$

Worked Example: Exercise 42

Calculate the volume of a 0.05 $\text{mol dm}^{-3}$ solution that contains 0.1 mol of solute.

$\text{Volume} = \frac{\text{amount}}{\text{concentration}} = \frac{0.1\text{ mol}}{0.05\text{ mol dm}^{-3}} = 2\text{ dm}^3$.

Saturated Solutions and Solubility

A saturated solution is defined as a solution in which no more solute can be dissolved at a given temperature. Solubility is the measure of the maximum mass of a substance that will dissolve in a specific amount of solvent, usually expressed as grams of solute per 100 grams of solvent ($\text{g} / 100\text{ g } \text{H}_2\text{O}$).

Interpreting Solubility Data

Solubility varies with temperature. For most solids, solubility increases as temperature increases. However, for gases, solubility typically decreases as temperature rises.

Any point on the curves shown in the diagram above represents a saturated solution at that specific temperature. For example, if you cool a hot saturated solution, the solubility decreases and the excess solute will precipitate out of the solution as a solid.

Calculating Solubility and Precipitation

Worked Example: Exercise 47

What is the maximum mass of copper(II) chloride ($\text{CuCl}_2$) that will dissolve in 250 $\text{cm}^3$ of water at $20^{\circ}\text{C}$? (Solubility is 73 g per 100 g $\text{H}_2\text{O}$, and the density of water is 1.0 $\text{g cm}^{-3}$).

1. Mass of water: $250\text{ cm}^3 \times 1.0\text{ g cm}^{-3} = 250\text{ g}$.

2. Calculate mass of solute: $\frac{73\text{ g}}{100\text{ g water}} \times 250\text{ g water} = 182.5\text{ g}$.

Worked Example: Exercise 48

Using the solubility curve below, what mass of copper(II) sulfate will precipitate if a saturated solution made with 50 g of water at $74^{\circ}\text{C}$ is cooled to $24^{\circ}\text{C}$?

1. At $74^{\circ}\text{C}$, solubility is 50 g per 100 g water. For 50 g water, the mass dissolved is $\frac{50}{2} = 25\text{ g}$.

2. At $24^{\circ}\text{C}$, solubility is 22 g per 100 g water. For 50 g water, the mass that can remain dissolved is $\frac{22}{2} = 11\text{ g}$.

3. Mass precipitated: $25\text{ g} - 11\text{ g} = 14\text{ g}$.