Chemistry · Chapter 11

Solutions and Colloids

Stir a spoonful of salt into water and the salt seems to vanish. Nothing was destroyed. The salt is still in the glass, split into ions and wrapped in water. That kind of mixture is a solution, and solutions run most of chemistry. Blood, seawater, soda, eye drops and brass are all solutions. This chapter shows you why some things dissolve and some refuse. You will follow the energy that dissolving costs and pays back. You will measure how much solute a mixture holds, five different ways. Then you will watch a pinch of solute move the freezing point and the boiling point of water.

By the end of this chapter you can

Describe dissolving as a trade of old attractions for new ones, and name the forces involved.
Use “like dissolves like” to predict whether a solute and a solvent will mix.
Add the three energy steps of dissolving to get ΔH_soln, and say what its sign means.
Work out molarity, molality, mole fraction, percent by mass and ppm, and convert between them.
Read a solubility curve, and use Henry's law to find how much gas dissolves at a given pressure.
Find the new freezing point, boiling point and osmotic pressure of a solution, using the van 't Hoff factor.
Tell a colloid apart from a true solution and a suspension, and explain the Tyndall effect.

Section 1The dissolving process

What counts as a solution

A solution is a mixture that looks the same all the way through. Take a drop from the top and a drop from the bottom and you get the same thing. The part present in the largest amount is the solvent. Anything dissolved in it is a solute. When water is the solvent we call the mixture aqueous and write (aq) after the formula.

Solutions are not always liquid. Air is a gas solution of oxygen and argon in nitrogen. Brass is a solid solution of zinc in copper. Water solutions matter most in a lab, so water is where we will spend our time.

Old attractions traded for new ones

Picture a grain of salt sitting in water. Inside the grain, Na⁺ and Cl^- ions hold each other hard. In the water, molecules hold each other through hydrogen bonds. For the salt to dissolve, both of those grips must loosen. In return, water molecules get to grab the ions.

That trade is the whole story of dissolving. It happens when the new attractions pay for the ones that broke. A water molecule has a slightly negative end at oxygen and a slightly positive end at each hydrogen. So the oxygen end swings toward Na⁺, and the hydrogen ends swing toward Cl^-. This pull between an ion and a polar molecule is an ion-dipole force.

Figure 11.1 Water molecules crowd the crystal face. The oxygen end turns toward Na⁺ and the hydrogen ends turn toward Cl^-. Each freed ion leaves with a shell of water around it.

The shell of solvent that forms around each dissolved particle is called solvation. When the solvent is water, the same thing gets its own name, hydration. A hydrated Na⁺ ion moves through the liquid with about six water molecules facing it oxygen-first. Those water molecules are not bonded to the ion. They swap places with others many times a second.

Like dissolves like

One short rule predicts most of what you will see at the bench. Like dissolves like means that polar solvents dissolve polar and ionic solutes, and nonpolar solvents dissolve nonpolar solutes.

Water is polar, so it handles salt and sugar well. Sugar has many O-H groups, and each one can hydrogen bond to water. Oil is nonpolar and is held together only by weak dispersion forces. To make room for oil, water would have to give up strong hydrogen bonds and would get very little back. So oil and water stay in layers. Hexane is nonpolar too, and it dissolves oil with ease.

Two liquids that mix in any amount are miscible. Water and ethanol are miscible, because the O-H group on ethanol hydrogen bonds to water. Water and oil are immiscible. Keep in mind that the rule is a guide, not a law. Some polar solutes still fail in water because their own lattice is simply too strong.

Key points

A solution is one phase: solvent plus solute, even all the way through.
Dissolving trades solute-solute and solvent-solvent attractions for solute-solvent attractions.
Ion-dipole forces hold hydrated ions. Hydrogen bonds hold sugar in water.
Like dissolves like. Polar goes with polar and ionic. Nonpolar goes with nonpolar.

Worked example · counting the ions that come free

You dissolve 0.250 mol of CaCl₂ in water. How many moles of ions end up in the solution?

Step 1 · write what the solid doesCaCl₂(s) → Ca²⁺(aq) + 2 Cl^-(aq)

Step 2 · read the ratioOne mole of the solid gives 1 mol of Ca²⁺ plus 2 mol of Cl^-. That is 3 mol of ions per mole of solid.

Step 3 · scale it0.250 mol CaCl₂ × (3 mol ions / 1 mol CaCl₂) = 0.750 mol ions

Answer0.750 mol of ions, made of 0.250 mol Ca²⁺ and 0.500 mol Cl^-. The 2+ charge on calcium pulls water harder than Na⁺ does, so it carries a tighter shell. Hold on to this count. Section 5 shows that the number of free particles, not their name, sets the freezing point.

Worth knowing. Dissolving is not melting and it is not a reaction. The ions in salt water are the same ions that sat in the crystal. Boil the water away and the crystal comes back.

Interactive Watch a solid dissolve and see like dissolve like. Open full ↑

Section 2Energy of solution formation

Three steps in your head, one event in the beaker

In the beaker, dissolving happens all at once. To track the energy, though, it helps to break the event into three imagined steps. The steps are a bookkeeping trick. Enthalpy does not care about the path, so any path that starts and ends in the right place gives the right total.

Step 1 pulls the solute apart. Its particles must be separated from each other, and that always costs energy. So ΔH₁ is positive. For an ionic solid, step 1 is the lattice energy paid in reverse.

Step 2 pushes the solvent apart to open a space. Water molecules give up some hydrogen bonds to make the cavity. This costs energy too, so ΔH₂ is positive.

Step 3 lets solute and solvent reach each other. New attractions form, so energy comes back out. ΔH₃ is negative. Add all three and you have the enthalpy of solution.

ΔH_soln = ΔH₁ + ΔH₂ + ΔH₃

Figure 11.2 Two uphill steps and one downhill step for NaCl. Steps 2 and 3 are measured as one value, the hydration enthalpy of -784 kJ/mol. The split shown here is an estimate of how much of that is cavity work.

Why the sign is hard to guess

For a salt in water, steps 2 and 3 are measured as one number. That number is the hydration enthalpy, the heat released when free gas-phase ions are surrounded by water. So the total is a race between two giants.

Breaking one mole of the NaCl lattice into free ions costs +787 kJ/mol. Hydrating that mole of Na⁺ and Cl^- gives back about -784 kJ/mol. What is left over is only about +3 kJ/mol. The measured value for NaCl is +3.9 kJ/mol. Both big numbers carry a few kJ of doubt, so a small leftover like this is always rough.

Tiny gaps decide the sign. NH₄NO₃ lands at +25.7 kJ/mol and chills the water, which is how an instant cold pack works. CaCl₂ lands at -81.3 kJ/mol and warms the water, which is how a hot pack works. Both are salts in water. The sign flips on a difference of a percent or two.

A positive value does not block dissolving

NH₄NO₃ dissolves freely even though it soaks up heat. Energy is only half the vote. Spreading solute through solvent also raises the disorder of the mixture, a quantity called entropy. That gain can carry the process on its own. Chapter 16 puts real numbers on the balance of the two.

Key points

ΔH_soln = ΔH₁ + ΔH₂ + ΔH₃. The first two are uphill, the third is downhill.
For salts, steps 2 and 3 together are the hydration enthalpy.
ΔH_soln is a small difference of two large numbers, so its sign is hard to predict.
Negative ΔH_soln warms the mixture. Positive ΔH_soln cools it.
A positive ΔH_soln can still dissolve, because entropy rises when solute spreads out.

Worked example · how cold does a cold pack get?

You stir 25.0 g of NH₄NO₃ into 150.0 g of water at 22.0 °C. For this salt ΔH_soln = +25.7 kJ/mol. Take the specific heat of the mixture as 4.18 J/g·°C. What is the final temperature?

Step 1 · moles of saltMolar mass of NH₄NO₃ = 14.01(2) + 1.008(4) + 16.00(3) = 80.05 g/mol
n = 25.0 g ÷ 80.05 g/mol = 0.3123 mol

Step 2 · heat taken inq = 0.3123 mol × 25.7 kJ/mol = 8.03 kJ = 8030 J

Step 3 · where that heat comes fromThe mixture itself gives it up. Its mass is 25.0 g + 150.0 g = 175.0 g.
ΔT = -q ÷ (m × c) = -8030 J ÷ (175.0 g × 4.18 J/g·°C)
ΔT = -8030 J ÷ 731.5 J/°C = -11.0 °C

AnswerThe mixture falls to 22.0 - 11.0 = 11.0 °C. Check the units: joules divided by joules per degree leaves degrees, which is what we want. The minus sign is right, because heat flowing into the dissolving salt has to leave the water.

Interactive Add up the three energy steps of dissolving. Open full ↑

Section 3Concentration units

Molarity: per liter of solution

Concentration says how crowded a solution is. The unit you pick depends on the job. Molarity (M) is moles of solute per liter of solution.

M = mol solute ÷ L solution

Molarity is the lab favorite, because you measure volume with a pipette in seconds. It has one flaw. Volume grows a little when a liquid warms, so the molarity of a bottle drifts as the room heats up.

Molality: per kilogram of solvent

Molality (m) is moles of solute per kilogram of solvent. Note the word solvent, not solution.

m = mol solute ÷ kg solvent

Mass never changes with temperature, so molality holds steady. That is why the freezing point and boiling point math in Section 5 uses molality and not molarity.

Figure 11.3 Molarity divides by the volume of the whole solution. Molality divides by the mass of solvent only. Density is the bridge between them.

Mole fraction, percent by mass and ppm

The mole fraction of a part is its moles divided by the total moles of everything present. Mole fractions have no units, and all of them in a sample add up to 1.

X_A = n_A ÷ n_total

Percent by mass is the mass of solute divided by the mass of the whole solution, times 100. For traces, scale up instead of down. Parts per million (ppm) uses a factor of 10⁶, and parts per billion uses 10⁹.

percent by mass = (mass solute ÷ mass solution) × 100 ppm = (mass solute ÷ mass solution) × 10⁶

For dilute water solutions there is a handy shortcut. A liter of such a solution weighs close to 1000 g, so 1 ppm is about 1 mg per liter. Drinking water limits are usually written this way.

Crossing between units

Percent, molality and mole fraction all run on mass, so you can swap among them with molar masses alone. To reach molarity you also need density, because molarity needs a volume. The trick is to choose a sample size that makes the numbers easy. For a percent, pick 100.0 g of solution.

Key points

Molarity is per liter of solution. Molality is per kilogram of solvent.
Molality and mass percent ignore temperature. Molarity does not.
Mole fraction is unitless, and the fractions in a sample sum to 1.
In dilute water solutions, 1 ppm is roughly 1 mg per liter.
You cannot reach molarity from a mass-based unit without the density.

Worked example · one solution in five units

A brine is 12.0% NaCl by mass. Its density is 1.086 g/mL. Find the molarity, the molality, the mole fraction of NaCl, and the concentration in ppm.

Step 1 · pick a sampleTake 100.0 g of solution. Then it holds 12.0 g of NaCl and 100.0 - 12.0 = 88.0 g of water.

Step 2 · moles of each partn(NaCl) = 12.0 g ÷ 58.44 g/mol = 0.2053 mol
n(H₂O) = 88.0 g ÷ 18.02 g/mol = 4.883 mol

Step 3 · molarity needs a volumeV = 100.0 g ÷ 1.086 g/mL = 92.08 mL = 0.09208 L
M = 0.2053 mol ÷ 0.09208 L = 2.23 M

Step 4 · molality uses the water massm = 0.2053 mol ÷ 0.0880 kg = 2.33 m

Step 5 · mole fractionX(NaCl) = 0.2053 ÷ (0.2053 + 4.883) = 0.2053 ÷ 5.088 = 0.0404

Step 6 · ppm(12.0 g ÷ 100.0 g) × 10⁶ = 120 000 ppm

Answer2.23 M, 2.33 m, X = 0.0404, 12.0% by mass, 120 000 ppm. Molality beats molarity here, as it always does for a solution denser than water. The water mass, 0.0880 kg, is smaller than the solution volume in liters, 0.09208 L, so the same moles get divided by less.

Worth knowing. Molality is written with a lower case m and molarity with a capital M. In handwriting that is a thin line to lean on, so label your work with words as well as symbols.

Interactive Convert molarity, molality, mole fraction, percent and ppm. Open full ↑

Section 4Solubility, temperature and pressure

Saturated, unsaturated and supersaturated

Solubility is the most solute that will dissolve in a fixed amount of solvent at a fixed temperature. A saturated solution holds exactly that much. Extra solid sits on the bottom, and it is not idle. Particles leave the solid and rejoin it at the same rate, so the amount in solution holds still.

Below the limit the solution is unsaturated and will take more. Careful cooling can leave a solution holding more than its limit. That state is supersaturated, and it is fragile. Drop in one seed crystal and the extra crashes out in seconds. A hand warmer full of sodium acetate works on exactly this trick.

Temperature: solids go both ways

Most solids dissolve better in hot water. KNO₃ climbs from 13.3 g per 100 g of water at 0 °C to 246 g at 100 °C. Some barely move. NaCl only goes from 35.7 g to 39.2 g over the same range. A few run backward. Ce₂(SO₄)₃ falls from 21.4 g at 0 °C to under 2 g at 100 °C.

There is no shortcut rule for solids. You read a solubility curve, which is a plot of solubility against temperature built from measured points.

Figure 11.4 Measured solubility of five salts in water. Points are table values at 0, 20, 40, 60, 80 and 100 °C, joined by straight lines.

Gases are different, and they are consistent. Warm water holds less dissolved gas. Set a glass of cold tap water on the counter and bubbles form on the inside wall as it warms. The same rule has teeth outside the lab. A river warmed by a power plant carries less oxygen, and fish feel it first.

Pressure and Henry's law

Pressure hardly touches the solubility of solids and liquids. For a gas it matters a great deal. Push harder on the gas above a liquid and more of it is forced in. Henry's law puts that in one line.

C = k_H P

Here C is the dissolved concentration, P is the partial pressure of that gas above the liquid, and k_H is the Henry's law constant for that gas in that solvent at that temperature. The law works well for gases that stay as molecules in the liquid and do not react with it.

Key points

A saturated solution is at balance: solid dissolves and re-forms at equal rates.
A supersaturated solution holds more than its limit and dumps the excess when seeded.
Solids mostly dissolve better hot, but you must read the curve for each one.
Gases always dissolve less in warmer liquid.
Henry's law, C = k_HP, works for gas solubility, not for solids.

Worked example · crystals from a cooling solution

You saturate 200.0 g of water with KNO₃ at 60 °C, then cool it slowly to 20 °C. From the curve, solubility is 110 g per 100 g of water at 60 °C and 31.6 g per 100 g at 20 °C. How much solid appears?

Step 1 · what 200.0 g of water holds when hot110 g / 100 g water × 200.0 g water = 220 g KNO₃

Step 2 · what it holds when cool31.6 g / 100 g water × 200.0 g water = 63.2 g KNO₃

Step 3 · subtract220 g - 63.2 g = 157 g

AnswerAbout 157 g of KNO₃ crystals. The grams of water cancel in both steps, leaving grams of salt. This is recrystallization, the standard way to clean up a solid product.

Worked example · oxygen in a lake, and gas in a soda

At 25 °C, k_H for O₂ in water is 1.3 × 10^-3 mol/L·atm. Air is 21% O₂ by volume. How much O₂ dissolves from air at 1.00 atm?

Step 1 · partial pressure of oxygenP = 0.21 × 1.00 atm = 0.21 atm

Step 2 · apply Henry's lawC = (1.3 × 10^-3 mol/L·atm)(0.21 atm) = 2.7 × 10^-4 mol/L

Step 3 · put it in mg/L2.7 × 10^-4 mol/L × 32.00 g/mol = 8.7 × 10^-3 g/L = 8.7 mg/L

That is close to the 8 to 9 mg/L a meter reads in cool, air-saturated water. Now the soda. For CO₂, k_H = 3.4 × 10^-2 mol/L·atm at 25 °C.

Step 4 · sealed can at 2.5 atm of CO₂C = (3.4 × 10^-2)(2.5) = 0.085 mol/L

Step 5 · opened can, where air holds only 3.9 × 10^-4 atm of CO₂C = (3.4 × 10^-2)(3.9 × 10^-4) = 1.3 × 10^-5 mol/L

AnswerOpening the can cuts the gas it can hold by a factor of 0.085 ÷ 1.3 × 10^-5, which is about 6500. The drink cannot lose that much in an instant, so it fizzes for a while instead.

Interactive Read solubility curves and apply Henry's law. Open full ↑

Section 5Colligative properties

Properties that count heads, not names

A colligative property depends only on how many solute particles are in the solvent. It does not care what they are. One mole of sugar and one mole of urea shift water's freezing point by the same amount. One mole of NaCl shifts it twice as far, because it breaks into two ions.

Vapor pressure lowering

Solute particles sit at the surface and take up room that water molecules would have used to escape. Fewer escape, so the vapor pressure falls. Raoult's law states the size of the drop.

P_solution = X_solvent × P°_solvent

Here P° is the vapor pressure of the pure solvent. Pure water at 25 °C is 23.8 torr. A solution with X_water = 0.9091 reads 0.9091 × 23.8 torr = 21.6 torr. Lower vapor pressure is the root cause of the next two effects.

Boiling point up, freezing point down

A liquid boils when its vapor pressure reaches the outside pressure. Lower the vapor pressure and you must heat the liquid further to get there. That is boiling point elevation. Solute also gets in the way of ice forming, so the liquid must be chilled further before it freezes. That is freezing point depression.

ΔT_b = i K_b m ΔT_f = i K_f m

The constants belong to the solvent, not the solute. For water, K_b = 0.512 °C/m and K_f = 1.86 °C/m. Notice that K_f is more than three times larger. That is why salting a road is worth doing and salting a pot of water barely helps it boil faster.

Figure 11.5 One molal NaCl in water. The freezing point drops 3.72 °C and the boiling point climbs 1.02 °C, so the liquid range widens by 4.74 °C.

The van 't Hoff factor

The van 't Hoff factor i is the number of particles one formula unit releases. Sucrose and other nonelectrolytes give i = 1. NaCl gives 2, CaCl₂ gives 3, and FeCl₃ gives 4. Real measurements fall a bit short of those whole numbers, because oppositely charged ions pair up for brief moments and travel as one particle. In dilute solution, measured i for NaCl is about 1.9, and for MgSO₄ it is near 1.3. The gap widens as the solution gets more crowded.

Osmotic pressure

Put a solution on one side of a membrane that passes water but not solute. Water moves toward the crowded side. That flow is osmosis. The pressure you must apply to stop it is the osmotic pressure, written Π.

Π = i M R T

Use R = 0.08206 L·atm/mol·K and T in kelvin. Osmotic pressure grows fast, which makes it the best of the four for measuring the molar mass of something rare. It also explains why a plant wilts in salty soil, and why saline for an IV drip is matched to blood at about 0.154 M NaCl.

Key points

Colligative properties count dissolved particles, whatever they are.
Raoult's law: the solvent's vapor pressure falls in step with its mole fraction.
ΔT_b = iK_bm and ΔT_f = iK_fm, with molality, not molarity.
Water: K_b = 0.512 °C/m, K_f = 1.86 °C/m.
i is 1 for a nonelectrolyte and equals the ions per formula unit for a salt, a little less in practice.
Π = iMRT, and it is the most sensitive of the four effects.

Worked example · antifreeze, then road salt

You mix 62.0 g of ethylene glycol, C₂H₆O₂, 62.07 g/mol, into 250.0 g of water. Find the new freezing and boiling points. Glycol is a nonelectrolyte, so i = 1.

Step 1 · moles and molalityn = 62.0 g ÷ 62.07 g/mol = 0.999 mol
m = 0.999 mol ÷ 0.2500 kg = 3.996 m

Step 2 · freezing pointΔT_f = (1)(1.86 °C/m)(3.996 m) = 7.43 °C
freezing point = 0.00 - 7.43 = -7.43 °C

Step 3 · boiling pointΔT_b = (1)(0.512 °C/m)(3.996 m) = 2.05 °C
boiling point = 100.00 + 2.05 = 102.05 °C

Now repeat with road salt. Use 5.85 g of NaCl, 58.44 g/mol, in the same 250.0 g of water, with i = 2.

Step 4 · moles and molalityn = 5.85 g ÷ 58.44 g/mol = 0.1001 mol
m = 0.1001 mol ÷ 0.2500 kg = 0.400 m

Step 5 · both shifts at onceΔT_f = (2)(1.86)(0.400) = 1.49 °C → -1.49 °C
ΔT_b = (2)(0.512)(0.400) = 0.410 °C → 100.41 °C

AnswerThe glycol sample freezes at -7.43 °C and boils at 102.05 °C. The salt sample freezes at -1.49 °C and boils at 100.41 °C. Check the units: °C/m times m leaves °C. Ten times less glycol than you might guess would do nothing, and one tenth the molality of salt still earns a 2× bonus from i.

Worked example · a molar mass from a freezing point, and the pressure in blood

You dissolve 1.50 g of an unknown nonelectrolyte in 75.0 g of water. The mixture freezes at -0.620 °C. What is the molar mass?

Step 1 · molality from the shiftm = ΔT_f ÷ (iK_f) = 0.620 °C ÷ (1 × 1.86 °C/m) = 0.3333 m

Step 2 · moles presentn = 0.3333 mol/kg × 0.0750 kg = 0.0250 mol

Step 3 · grams per moleM = 1.50 g ÷ 0.0250 mol = 60.0 g/mol

Next, the osmotic pressure of a saline drip: 0.154 M NaCl at body temperature, 37 °C, with i = 2.

Step 4 · convert the temperatureT = 37 + 273.15 = 310.15 K

Step 5 · apply Π = iMRTΠ = (2)(0.154 mol/L)(0.08206 L·atm/mol·K)(310.15 K) = 7.8 atm

Answer60.0 g/mol, and about 7.8 atm. Urea fits that molar mass. The 7.8 atm figure sits right on the measured osmotic pressure of blood, which is why this strength of saline leaves red cells unharmed.

Interactive Add solute and watch freezing point, boiling point and osmosis respond. Open full ↑

Section 6Colloids and suspensions

Three mixtures, sorted by particle size

Not every cloudy liquid is a solution. Sort mixtures by how big the scattered bits are. In a true solution the particles are single ions or molecules, under about 1 nm across. In a suspension such as muddy water, the bits are larger than about 1000 nm. They settle out and filter paper catches them.

Between those two sits the colloid. Its particles run from about 1 nm to 1000 nm. They are too small to settle and too large to slip through a membrane. Milk, fog, jelly, whipped cream and paint are all colloids. The scattered material is the dispersed phase and the material it floats in is the dispersion medium.

Colloids stay up for two reasons. Constant random kicks from solvent molecules, called Brownian motion, keep knocking the particles around. Many colloid particles also carry the same surface charge, so they push each other apart instead of clumping.

Figure 11.6 The same lamp aimed at three mixtures. Only the middle tube, the colloid, shows the beam as a bright streak.

The Tyndall effect

Shine a lamp through a true solution and the beam stays invisible from the side. The particles are far smaller than a wavelength of light, so they barely scatter it. Shine the same lamp through a colloid and the beam shows as a bright streak. This scattering by colloid-sized particles is the Tyndall effect. Sunbeams through dusty air and headlights in fog are the same effect outdoors. It is the quickest bench test for a colloid.

Emulsions and the agents that hold them

An emulsion is liquid droplets spread through another liquid that they do not mix with. Shake oil and water and you make one for a moment. The droplets then find each other, merge and split into layers again. An emulsifying agent stops that. Its molecules have a polar head and a nonpolar tail, so they coat each droplet with the tail inward and the head facing the water.

Egg yolk does this job in mayonnaise, using a molecule called lecithin. Soap does it when it lifts grease off a pan. Milk is an oil-in-water emulsion held by proteins. Heating, freezing or adding acid can strip the coating, and then the emulsion breaks.

Colloids come in other pairs too. A liquid spread in a gas is an aerosol, as in fog. A gas in a liquid is a foam, as in whipped cream. A solid in a liquid is a sol, as in paint, and a liquid trapped in a solid network is a gel, as in jelly.

Key points

Solution under 1 nm, colloid 1 to 1000 nm, suspension over 1000 nm.
Colloids do not settle. Suspensions do.
Only colloids show the Tyndall effect, so a lamp tells them apart in seconds.
Brownian motion and like surface charges keep colloid particles spread out.
An emulsion needs an emulsifying agent with a polar head and a nonpolar tail.

Worked example · why colloids have so much surface

Cut a cube of solid 1.00 cm on each edge into cubes 100 nm on each edge. How much surface area do you end up with?

Step 1 · one set of units100 nm = 1.00 × 10^-5 cm, and 1.00 cm = 1.00 × 10⁷ nm.

Step 2 · count the small cubesEach edge holds 1.00 × 10⁷ ÷ 100 = 1.00 × 10⁵ of them.
number = (1.00 × 10⁵)³ = 1.00 × 10¹⁵ cubes

Step 3 · area of one small cubeA = 6 × (1.00 × 10^-5 cm)² = 6.00 × 10^-10 cm²

Step 4 · total area(1.00 × 10¹⁵)(6.00 × 10^-10 cm²) = 6.00 × 10⁵ cm² = 60.0 m²

Answer60.0 m², up from 6.00 cm² for the whole cube. That is 100 000 times more surface, on the same mass. Surface is where a colloid does its work, which is why foams, inks and catalysts are built at this size.

Worth knowing. Add salt to a colloid and you can wreck it. The ions screen the surface charges, so the particles stop repelling and clump together. River silt drops out where fresh water meets the sea for exactly this reason, and that is how a delta is built.

Interactive Shine a light through solutions and colloids for the Tyndall effect. Open full ↑

The chapter in one breath

A solution is one even phase. Dissolving swaps old attractions for new ones, and like dissolves like.
Water hydrates ions with ion-dipole forces, turning its oxygen end to cations and its hydrogen ends to anions.
ΔH_soln is two uphill steps plus one downhill step. For salts it is a small leftover from two huge numbers.
A positive ΔH_soln still dissolves, because spreading out raises entropy.
Molarity is per liter of solution and molality is per kilogram of solvent. Density bridges them.
Solids follow their own solubility curve with temperature. Gases always dissolve less when warm.
Henry's law, C = k_HP, sets how much gas dissolves at a given partial pressure.
Solute lowers vapor pressure, so ΔT_b = iK_bm and ΔT_f = iK_fm. Water: 0.512 and 1.86 °C/m.
Colloid particles span 1 to 1000 nm, never settle, and show the Tyndall effect.

ReferenceWords worth knowing

These are the words this chapter leans on. Say each one out loud with its meaning before you try the questions below.

boiling point elevation: The rise in a solvent's boiling point when solute is added. ΔT_b = iK_bm, and K_b for water is 0.512 °C/m.
colligative property: A property set by how many solute particles are present, not by what they are. Vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure.
colloid: A mixture whose dispersed particles measure about 1 nm to 1000 nm. It does not settle and it scatters light.
emulsion: A colloid of liquid droplets in another liquid, held apart by an emulsifying agent with a polar head and a nonpolar tail.
freezing point depression: The drop in a solvent's freezing point when solute is added. ΔT_f = iK_fm, and K_f for water is 1.86 °C/m.
Henry's law: C = k_HP. The amount of a gas dissolved in a liquid is proportional to that gas's partial pressure above the liquid.
hydration: Solvation when the solvent is water. A shell of water molecules surrounds each dissolved particle.
ion-dipole force: The attraction between an ion and the charged end of a polar molecule. It is what holds a hydrated ion together.
miscible: Able to mix in any proportion, said of two liquids. Liquids that form layers instead are immiscible.
molality: Moles of solute per kilogram of solvent, symbol m. It does not change when the sample warms.
molarity: Moles of solute per liter of solution, symbol M. Easy to measure, but it shifts slightly with temperature.
mole fraction: The moles of one part divided by the total moles present. It has no units, and all the fractions in a sample add to 1.
osmosis: The flow of solvent through a membrane that blocks solute, running toward the more crowded side.
osmotic pressure: The pressure needed to stop osmosis, Π = iMRT. The most sensitive of the colligative properties.
parts per million: Mass of solute divided by mass of solution, times 10⁶. In dilute water solutions 1 ppm is about 1 mg per liter.
Raoult's law: P_solution = X_solventP°_solvent. The solvent's vapor pressure falls in step with its mole fraction.
saturated solution: A solution holding all the solute it can at that temperature, with dissolving and re-forming in balance. Holding more than that is supersaturated and unstable.
solvation: The wrapping of solvent molecules around a dissolved solute particle.
Tyndall effect: The scattering of a light beam by colloid-sized particles, which makes the beam visible from the side.
van 't Hoff factor: Symbol i, the number of particles one formula unit of solute releases. It is 1 for a nonelectrolyte, 2 for NaCl, and a little below the whole number in practice.

Check yourself

1. Iodine, I₂, barely dissolves in water but dissolves well in hexane. Explain in terms of forces.

I₂ is nonpolar. Hexane is nonpolar too, so only weak dispersion forces have to be traded on both sides. Water would have to break strong hydrogen bonds and would get almost nothing back, so like dissolves like predicts the split.

2. What is the molarity of a solution made by dissolving 4.25 g of NaCl in enough water to reach 250.0 mL?

0.291 M n = 4.25 g ÷ 58.44 g/mol = 0.0727 mol. V = 250.0 mL = 0.2500 L. M = 0.0727 mol ÷ 0.2500 L = 0.291 mol/L.

3. A cold pack uses NH₄NO₃ with ΔH_soln = +25.7 kJ/mol. Why does the pack feel cold, and why does the salt dissolve at all?

It absorbs heat, and entropy pays the bill. A positive ΔH_soln means the dissolving salt pulls heat from the water, so the mixture cools. Spreading the ions through the water raises entropy enough to drive the change even though it costs energy.

4. You mix 90.0 g of ethylene glycol, 62.07 g/mol, into 500.0 g of water. What is the freezing point? Glycol is a nonelectrolyte.

-5.39 °C n = 90.0 g ÷ 62.07 g/mol = 1.450 mol. m = 1.450 mol ÷ 0.5000 kg = 2.900 m. ΔT_f = (1)(1.86 °C/m)(2.900 m) = 5.39 °C, so the new freezing point is 0.00 - 5.39 = -5.39 °C.

5. Find the boiling point of a solution of 11.7 g of NaCl in 200.0 g of water. Use i = 2.

101.02 °C n = 11.7 g ÷ 58.44 g/mol = 0.200 mol. m = 0.200 mol ÷ 0.2000 kg = 1.00 m. ΔT_b = (2)(0.512 °C/m)(1.00 m) = 1.02 °C. Boiling point = 100.00 + 1.02 = 101.02 °C.

6. At 25 °C, k_H for N₂ in water is 6.0 × 10^-4 mol/L·atm. Air at 1.00 atm is 78% N₂. How much N₂ dissolves?

4.7 × 10^-4 mol/L P(N₂) = 0.78 × 1.00 atm = 0.78 atm. C = k_HP = (6.0 × 10^-4 mol/L·atm)(0.78 atm) = 4.7 × 10^-4 mol/L. In mass terms that is about 13 mg/L.

7. Seawater is about 3.5% salt by mass. Write that in ppm.

35 000 ppm ppm = (mass solute ÷ mass solution) × 10⁶. A mass percent of 3.5% is 3.5 g in 100 g, so (3.5 ÷ 100) × 10⁶ = 3.5 × 10⁴ = 35 000 ppm.

8. What is the osmotic pressure of a 0.0150 M glucose solution at 25 °C? Glucose is a nonelectrolyte.

0.367 atm T = 25 + 273.15 = 298.15 K. Π = iMRT = (1)(0.0150 mol/L)(0.08206 L·atm/mol·K)(298.15 K) = 0.367 atm. Units: mol/L times L·atm/mol·K times K leaves atm.

9. A 2.40 g sample of an unknown nonelectrolyte in 60.0 g of water freezes at -1.24 °C. What is its molar mass?

60.0 g/mol m = ΔT_f ÷ (iK_f) = 1.24 °C ÷ (1 × 1.86 °C/m) = 0.667 m. n = 0.667 mol/kg × 0.0600 kg = 0.0400 mol. M = 2.40 g ÷ 0.0400 mol = 60.0 g/mol.

10. A clear solution sits over a pile of undissolved crystals and the pile never shrinks. Is the solution saturated, unsaturated or supersaturated?

Saturated. The solid is still dissolving, but solute is rejoining the crystals at the same rate. Those equal rates are what a saturated solution means. A supersaturated solution holds extra solute and has no solid left sitting in it.

11. You aim a flashlight at two clear glasses. One beam is invisible from the side and the other shows as a bright streak. What is in each glass?

Solution, then colloid. Particles in a true solution are far smaller than a wavelength of light and scatter very little. Colloid particles, 1 nm to 1000 nm across, scatter enough to make the beam show. That is the Tyndall effect.

12. Fish struggle in a pond during a heat wave even when the water is not polluted. Explain using gas solubility.

Warm water holds less oxygen. Gas solubility always falls as temperature rises, because warmer dissolved molecules have the energy to break away and leave. Henry's law also applies, so the partial pressure of oxygen in the air above the pond sets the ceiling.

Previous← Chapter 10 NextChapter 12 →