A solubility value tells you how much solute can dissolve in a stated mass of solvent at a stated temperature. The most common calculation error is using the whole solution mass where the data refer to water alone. A second is assuming that every solid added at the start actually dissolved.
Here you will use a fictional salt S and an original illustrative dataset. S is an exercise label, not a real salt formula. All calculations are paper models. For the fictional-S solubility calculations, assume equilibrium is reached, no water evaporates, no material is lost, no chemical reaction occurs, and any crystals are anhydrous S, so crystallisation does not remove water into a hydrate.
Read the axes and the unit together
| Temperature (°C) | Solubility (g S per 100 g water) |
|---|---|
| 20 | 20 |
| 30 | 28 |
| 40 | 38 |
| 50 | 50 |
| 60 | 64 |
| 70 | 80 |
Complete graph specification: Temperature in °C is the horizontal axis and solubility in grams of S per 100 g water is the vertical axis. The ordered points are (20,20), (30,28), (40,38), (50,50), (60,64) and (70,80). Straight segments join consecutive points only within 20–70°C. These segments define the interpolation convention for this illustrative graph; they are not measured behaviour of a real salt.
At 60°C, the value 64 means that a saturated equilibrium solution can contain 64 g S for each 100 g water. Its total solution mass would be 164 g, not 100 g. A solution containing only 40 g S in that same water mass would be unsaturated at 60°C under the model.
Use capacity = solubility × water mass ÷ 100. For 60 g water at 60°C, the capacity is 64 × 60/100 = 38.4 g S. This is the maximum equilibrium dissolved mass for the stated water quantity, not necessarily the amount that was added.
Worked analysis 1: cool an initially saturated solution
A saturated solution at 60°C contains 60 g water. It is cooled to 20°C with no water loss, and crystallisation reaches equilibrium. Calculate the initial dissolved mass, final dissolved mass and crystals formed.
- Initial dissolved S = 64 × 60/100 = 38.4 g.
- Final dissolved capacity = 20 × 60/100 = 12.0 g.
- Crystals formed = 38.4 − 12.0 = 26.4 g.
Check conservation of solute: 26.4 g crystals + 12.0 g dissolved = 38.4 g initially dissolved. Check the complete mass: the original solution is 60 + 38.4 = 98.4 g; the final system is 72.0 g liquid solution plus 26.4 g crystals, again 98.4 g.
The difference between curve values is 64 − 20 = 44 g per 100 g water. It is not yet the crystal mass. Scaling 44 by 60/100 gives 26.4 g. Writing 44 g as the answer silently changes the problem to one containing 100 g water.
Worked analysis 2: the added solid exceeds hot solubility
Now 50.0 g S is introduced into 60 g water at 60°C. Equilibrium is reached. The whole mixture, including any undissolved solid, is then cooled to 20°C. Calculate each solid and dissolved mass carefully.
The hot liquid can dissolve only 38.4 g, so 50.0 − 38.4 = 11.6 g remains undissolved at 60°C. On cooling, the dissolved mass falls to 12.0 g. Thus 26.4 g crystallises from the liquid during cooling, but the total solid present at the end is 11.6 + 26.4 = 38.0 g.
These two answers describe different quantities. “New crystals from cooling” is 26.4 g. “All solid present at 20°C” is 38.0 g. A question that asks one cannot be answered with the other. The final solute check is 38.0 + 12.0 = 50.0 g.
If the initial 11.6 g undissolved material had instead been removed before cooling, the remaining liquid would start with only 38.4 g S. It would still form 26.4 g crystals, but the final vessel would contain no additional original solid. Track whether a separation step removes material from the system you are considering.
Worked analysis 3: a mass given for the whole solution
A 150 g saturated solution of S at 50°C cools to 20°C. How much S crystallises? The phrase “150 g solution” does not mean 150 g water.
At 50°C the saturated ratio is 50 g S to 100 g water, giving 150 g solution in total. The stated sample therefore contains 100 g water and 50 g dissolved S. At 20°C those 100 g water retain 20 g S, so 50 − 20 = 30 g crystallises.
The general recovery of water mass is: water mass = solution mass × 100/(100 + hot solubility). Here this is 150 × 100/150 = 100 g. This relation applies to a saturated solution with no separate undissolved solid, because the specified ratio must describe the entire weighed sample.
The incorrect calculation (50 − 20) × 150/100 = 45 g treats all 150 g as water. It therefore assumes an initial solute amount of 75 g, which contradicts the original 150 g saturated-solution mass. A composition check exposes the error before the final subtraction.
Worked analysis 4: not enough solute to reach saturation
A solution contains 18 g S and 60 g water at 60°C, with all the S dissolved. It cools to 40°C. The final capacity is 38 × 60/100 = 22.8 g, which exceeds the available 18 g. Therefore all 18 g remains dissolved and no crystals form.
Using the saturated-curve difference would give (64 − 38) × 0.60 = 15.6 g, but the starting solution was not saturated. The hot capacity tells you what could dissolve, not what is actually present. The problem explicitly supplied the initial dissolved amount.
If that same solution cools further to 20°C, the capacity becomes 12 g. It then forms 18 − 12 = 6 g crystals. A convenient general rule for these equilibrium cases is final dissolved mass = the smaller of available solute mass and final capacity. Final solid is the remaining solute, never a negative mass.
What an above-curve point does and does not mean
The interpretation depends on what the plotted quantity represents. If it represents total S added per 100 g water and equilibrium is established, a point above the solubility line means some S remains solid. If it represents S actually dissolved, an above-line value could describe a metastable supersaturated solution.
A dot alone cannot distinguish those situations. You need the definition of the vertical quantity and information about equilibrium or solid formation. Our tasks explicitly assume equilibrium and distinguish dissolved mass from total added mass, so that ambiguity is removed.
Likewise, the rising pattern for fictional S is not a rule that every solid becomes more soluble as temperature rises. Use the curve supplied for the substance. Do not extrapolate this six-point model beyond 20–70°C without further data.
Three practice tasks with solved answers
Practice 1: change the water quantity
A saturated S solution containing 75 g water cools from 70°C to 30°C. Find the initial dissolved mass, final dissolved mass and crystals formed.
Answer: Initially 80 × 0.75 = 60 g S dissolves. Finally 28 × 0.75 = 21 g remains dissolved. Crystals = 60 − 21 = 39 g. The solute check is 39 + 21 = 60 g. The water mass stays 75 g throughout.
Practice 2: distinguish new crystals from total solid
A mixture contains 30 g S and 40 g water at 50°C. After reaching equilibrium, the entire mixture cools to 20°C. Find the initial undissolved mass, newly crystallised mass and final total solid.
Answer: Hot capacity = 50 × 0.40 = 20 g, leaving 10 g initially undissolved. Cold capacity = 20 × 0.40 = 8 g. New crystals = 20 − 8 = 12 g. Final total solid = 10 + 12 = 22 g. Check: 22 g solid + 8 g dissolved = 30 g S.
Practice 3: interpolate inside the stated model
Use the straight segment between 40°C and 50°C. Estimate the solubility at 45°C, then find the crystals formed when a saturated solution containing 50 g water cools from 45°C to 20°C.
Answer: Halfway between 38 and 50 gives 44 g S per 100 g water. Initially 44 × 0.50 = 22 g is dissolved. Finally 20 × 0.50 = 10 g remains. Crystals = 12 g. This interpolation is defined by the exercise graph; a real experimental curve need not be exactly linear between measurements.
Choose a salt route from solubility and reaction evidence
Solubility data tell you where the product can be found; the chemical equation tells you what product can form. A preparation-method question needs both. These method choices and records are for paper analysis only. They are not instructions for handling acids, making salts or heating solutions, and no practical activity is needed.
| Target and supplied properties | Suitable route class | Separation logic |
|---|---|---|
| Soluble salt; suitable insoluble basic reactant available | Acid–insoluble-base neutralisation | Unreacted solid can be distinguished from the salt-bearing solution; the dissolved product is not the initial filter residue. |
| Soluble salt; both acid and alkali are soluble | Neutralisation with a titration-based amount decision | An excess of dissolved reactant cannot be removed by ordinary filtration. The reacting proportions must be established. |
| Sparingly soluble target; suitable soluble ion sources available | Precipitation | The target forms a solid fraction, while spectator ions remain mainly in solution. |
This is a choice among the stated options, not a claim that every salt has only one possible industrial route. A colour change is not required for every neutralisation. The retained titration calculations show how an amount decision is obtained, and the separation lesson explains why dissolved salt passes an ordinary filter.
Worked record: a soluble salt stays in the liquid
A fictional card names CuO as an insoluble basic solid and CuSO₄ as soluble under the supplied conditions. The reported reaction is CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l). A later record says the acid has been consumed, some black CuO remains and the liquid is blue. Which route fits, and where is the desired salt?
The acid–insoluble-base route fits the stated reactants. Unreacted CuO is the solid fraction; dissolved CuSO₄ is in the liquid fraction. The acid-consumed statement is supplied evidence, not something established solely by seeing leftover solid. Ordinary filtration could distinguish solid from solution in this model, but would not recover dissolved CuSO₄ as its residue. Crystal recovery belongs to a later solubility change. The curve above teaches that calculation for fictional anhydrous S; its constants must not be silently reused as real CuSO₄ data or a hydrate model.
Worked record: an insoluble salt is the new solid
Another card supplies CaCl₂(aq) and Na₂CO₃(aq), with enough of both ions for precipitation. It reports a white solid and gives CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaCl(aq). The suitable route class is precipitation. The desired CaCO₃ is now the solid fraction, the opposite product location from the soluble-salt case.
The net ionic equation is Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s): charge +2 − 2 becomes zero, while Ca, C and O atoms are conserved. Sodium and chloride remain spectator ions in the solution model. “A white solid formed” is the observation; CaCO₃ is its supported identity because the reactants and reaction model are supplied. Appearance alone would not distinguish every possible white salt.
Independent route decisions
- A sodium chloride target is soluble, and the only specified reactants are aqueous HCl and NaOH. Choose a route class, write its equation and explain why excess dissolved NaOH cannot simply be filtered away.
- A card uses the precipitation model above but labels the filtrate “pure solid CaCO₃”. Correct the location and the overclaim about purity.
- Return to the existing S curve. A saturated solution with 50 g water cools from 60°C to 20°C with no loss and anhydrous crystals. Calculate crystal mass and explain which assumption would fail for a hydrate.
Checked answers
1. Titration-based neutralisation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). All coefficients are one. Dissolved excess alkali remains in the liquid through ordinary filtration. 2. The CaCO₃ is the solid residue in this ideal separation; the filtrate contains dissolved ions. A real solid’s purity would need evidence about retained solution and other components. 3. Initially 64 × 0.50 = 32 g S; finally 20 × 0.50 = 10 g dissolved; crystals = 22 g. A hydrate incorporates water into its crystals, so the unchanged-water-mass bookkeeping would need revision.
Background: RSC, soluble-salt method rationale and OpenStax, precipitation reactions. Only explanatory chemistry is used; no source procedure is reproduced.
Frequently asked questions
What does 64 g per 100 g water mean?
At the stated temperature, that is the saturated mass ratio for fictional S. It describes 64 g solute with 100 g water, giving 164 g solution if no separate solid is included.
Can I use the total solution mass in the capacity formula?
No. The formula uses water mass. If a saturated solution mass is supplied, first use its solute-to-water ratio to find the water portion.
Why might cooling form no crystals?
An initially unsaturated solution may still contain less solute than the colder water can dissolve. Compare actual solute mass with final capacity before subtracting.
Is final solid always the same as new crystals?
No. If undissolved solid was already present and remained in the vessel, final solid includes that material plus any additional solute crystallised during cooling.
Does a point above the curve always mean undissolved solid?
No. It depends on whether the plotted quantity is total added solute or actually dissolved solute, and whether equilibrium has been reached. A supersaturated solution is another possibility.
May I extend this graph to 100°C?
Not reliably. The illustrative data and interpolation convention apply only from 20°C to 70°C. No values outside that range have been supplied.
What assumptions keep the water mass unchanged?
There is no evaporation or loss, and the model crystals are anhydrous S. A hydrate forming with water of crystallisation would need different bookkeeping.
Sources and further reading
The references below support the scientific background. The cases, calculations and illustrations in this lesson are original. External teaching guidance does not establish current KCSE marking rules or grade placement.
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