Solutions: concentration units, the dissolving process and solubility

Until now we have known solutions only through molarity, because it was useful in stoichiometry. This week we study the solution on its own. There are two questions: how much is dissolved? (concentration units and conversions among them) and why does it dissolve? (intermolecular forces, energy and solubility). Next week’s colligative properties need molality and mole fraction; this week you will learn to use them fluently.

Golden rules:

1. Molarity divides by the volume of the solution, molality by the mass of the solvent. The two units most often confused.
2. In conversion questions, assume a convenient amount. 1 L of solution, 1 kg of solvent or 100 g of solution.
3. Like dissolves like. Polar dissolves polar, nonpolar dissolves nonpolar.
4. The solubility of gases increases with pressure and decreases with temperature.

Concentration units

Example 1 (one solution, all the units). 25.0 g of KBr (119.00 g/mol) is dissolved in 175.0 g of water. The density of the solution is 1.15 g/mL.

n(KBr)       = 25.0 / 119.00          = 0.2101 mol
n(water)     = 175.0 / 18.015         = 9.714 mol
% mass       = 25.0 / 200.0 × 100     = 12.5 %
molality     = 0.2101 / 0.1750 kg     = 1.20 m
mole fraction = 0.2101 / (0.2101 + 9.714) = 0.0212
V(solution)  = 200.0 g / 1.15 g/mL    = 173.9 mL
molarity     = 0.2101 / 0.1739 L      = 1.21 M

In dilute aqueous solutions molarity and molality are close to each other, because 1 L of solution is about 1 kg of water. As the concentration increases, the difference grows.

Example 2 (ppm). 0.0040 g of mercury in 2.00 kg of water:

$\frac{0.0040\ \text{g}}{2000\ \text{g}}\times10^6 = 2.0\ \text{ppm} = 2.0\times10^3\ \text{ppb}$

In a dilute aqueous solution 1 L ≈ 1 kg, so 1 ppm ≈ 1 mg/L.

Conversion between units

All conversions rest on the same idea: find each quantity in the definition (moles of solute, mass of solvent, volume of solution) one by one. For this, assume a convenient amount.

Example 3 (from molarity to molality). 0.800 M NaCl, density 1.03 g/mL.

assume 1.000 L of solution
mass(solution) = 1000 mL × 1.03 g/mL     = 1030 g
NaCl           = 0.800 mol × 58.44 g/mol = 46.75 g
water          = 1030 − 46.75            = 983.3 g = 0.9833 kg
molality       = 0.800 / 0.9833          = 0.814 m

From molality to molarity do the reverse: assume 1.000 kg of solvent, add the mass of the solute, and find the volume by dividing the total mass by the density.

Example 4 (from molality to molarity). 2.00 m NaOH, density 1.07 g/mL.

assume 1.000 kg water → 2.00 mol NaOH = 2.00 × 39.997 = 80.0 g
mass(solution) = 1000 + 80.0  = 1080 g
V(solution)    = 1080 / 1.07  = 1009 mL = 1.009 L
molarity       = 2.00 / 1.009 = 1.98 M

From molality to mole fraction: 1 kg of water = 55.51 mol. $x = \frac{m}{m + 55.51}$ (if water is the solvent).

PITFALL · dividing by the mass of the solution in molality. The denominator of molality is the mass of the solvent. If you assumed 100 g of solution in a percent question, the water is not 100 g but 100 − (solute) g.

The dissolving process

Think of the dissolving of a substance in three steps (like Hess’s law):

1. Separate the solute particles from each other (solute-solute interactions are broken): endothermic.
2. Spread the solvent molecules apart to make room (solvent-solvent interactions are broken): endothermic.
3. Solute and solvent particles interact (solute-solvent interactions form): exothermic.

$\Delta H_{\text{soln}} = \Delta H_1 + \Delta H_2 + \Delta H_3$

• If the third step is larger than the first two, $\Delta H_{\text{soln}} < 0$: dissolving gives off heat (NaOH, $\mathrm{H_2SO_4}$ get very hot in water).
• If it is smaller, $\Delta H_{\text{soln}} > 0$: the solution gets cold ($\mathrm{NH_4NO_3}$ cold packs).
• Endothermic dissolving can also be spontaneous, because mixing increases the disorder (entropy). But if $\Delta H_{\text{soln}}$ is very positive (oil and water) the entropy cannot compensate for it, and no dissolving occurs.

When an ionic solid dissolves in water, each ion is surrounded by water molecules: hydration. This ion-dipole interaction provides the energy of step 3 and tries to pay for the lattice energy (step 1).

Like dissolves like

A long hydrocarbon chain reduces polarity: methanol, ethanol and propanol mix with water in any proportion; the solubility of butanol is limited; octanol is almost insoluble. The O–H end likes water, the chain likes hexane.

Solubility

Solubility is the largest amount of solute that can dissolve in a given amount of solvent at a given temperature (usually g / 100 g of water).

• Unsaturated: can take more solute.
• Saturated: in equilibrium with undissolved solid; the rates of dissolving and crystallization are equal.
• Supersaturated: contains more solute than the solubility, unstable. When a small crystal (a seed) is added, the excess comes out immediately. It is prepared by slowly cooling a hot saturated solution.

The effect of temperature

• The solubility of most solids increases with temperature ($\mathrm{KNO_3}$ a lot, NaCl very little). There are a few exceptions (like $\mathrm{Ce_2(SO_4)_3}$). This difference is the basis of purification by recrystallization.

Example 5 (crystallization from a solubility curve). Let the solubility of a salt be 110 g / 100 g of water at 60 °C and 32 g / 100 g of water at 20 °C. If a saturated solution made with 100 g of water at 60 °C is cooled to 20 °C, at most 32 g can remain in the water; $110 - 32 = 78$ g of salt separates as crystals. Since the impurities are present in small amounts, they stay in the solution; the filtered crystals are purer.

• The solubility of gases in water decreases with temperature. In heated water, bubbles form before boiling: dissolved air comes out.

The effect of pressure: Henry’s law

The solubility of solids and liquids is hardly affected by pressure. That of gases is directly proportional to the partial pressure:

$C_{\text{gas}} = k_H \cdot P_{\text{gas}}$

$k_H$ is a constant that depends on the gas, the solvent and the temperature (M/atm). $P_{\text{gas}}$ is the partial pressure of that gas (Dalton’s law).

Example 6. Oxygen dissolved in water in equilibrium with air ($P_{\mathrm{O_2}} = 0.21$ atm, $k_H = 1.3\times10^{-3}$ M/atm):

$C = 1.3\times10^{-3} \times 0.21 = 2.7\times10^{-4}\ \text{M} \;\Rightarrow\; 2.73\times10^{-4} \times 32.00 \times 1000 = 8.7\ \text{mg/L}$

You can also use Henry’s law as a ratio for two situations: $\frac{C_1}{P_1} = \frac{C_2}{P_2}$. If you raise the pressure by a factor of 2.5, the solubility also rises by a factor of 2.5. Divers getting the bends is related to this: at depth more $\mathrm{N_2}$ dissolves in the blood, and when they rise quickly bubbles form.

Common mistakes