Solubility at a Stated Temperature

Introduction

A solubility number is meaningful only with its conditions and basis. “36 g per 100 g water at 25 °C” is a different statement from “36 g per 100 g solution,” and a change in temperature can change the value. Read the whole statement before predicting what will remain dissolved.

Core explanation

Solubility is the equilibrium maximum dissolved amount under specified conditions when excess solute is present. A common school unit for a solid in water is grams of solute per 100 g water at a stated temperature. If a fictional salt has solubility 40 g per 100 g water at 20 °C, then 50 g water can hold 20 g dissolved salt at equilibrium, assuming the same solvent composition and no reaction. Adding 15 g can produce an unsaturated solution if it dissolves fully; adding 30 g leaves up to 10 g as a separate solid at equilibrium.

The denominator is solvent mass, not solution mass. In the example, 20 g dissolved solute with 50 g water makes 70 g saturated liquid solution, excluding undissolved material. Its solute mass fraction is 20/70 ≈ 0.286. Reporting “40 g per 100 g water” as 40% by mass of solution would be incorrect; at that limit the solution mass is 140 g, so mass fraction is 40/140 ≈ 0.286. This conversion is a frequent exam trap.

Saturation is dynamic. At equilibrium, particles can still leave a crystal surface while others return, with no net change in dissolved concentration. It does not mean that molecular motion stops. If temperature changes, a different equilibrium solubility may apply, and crystals may dissolve or form. The direction and size of the change require data for the particular solute. Pressure can be important for gases; for many solid-in-liquid school problems, temperature and solvent amount are emphasized.

Solubility belongs to a defined solute and solvent. The presence of a common ion, another solute or a mixed solvent can change the equilibrium for some systems. A numerical table measured for pure water may not describe a complex mixture exactly. When a problem gives an idealized solubility curve, use its specified assumptions and avoid adding unsupported effects.

The mass added is not always the mass dissolved. A clear portion of liquid above excess crystals contains dissolved solute at saturation, while the whole beaker includes an additional solid phase. Filtering off excess solid can isolate the saturated liquid without changing its equilibrium concentration immediately at the same conditions. The total sample mass and solution mass must be distinguished in calculations.

Step-by-step reasoning

1. Read the solubility unit, solute, solvent and temperature. 2. Scale the given limit to the actual solvent amount. 3. Compare that capacity with the amount of solute added. 4. Take the smaller amount as dissolved at equilibrium; the remainder is excess solid. 5. If asked for mass fraction, divide dissolved mass by the liquid solution mass.

Visual explanation

Draw a balance scale labeled “40 g solute per 100 g water at 20 °C.” For 50 g water, halve both figures to show a 20 g capacity. Place 30 g added solute beside it; shade 20 g as dissolved and 10 g as solid at the bottom.

Real-world analogy

A bus can carry only a stated number of passengers. A smaller bus has proportionally fewer spaces; extra passengers remain outside. The analogy captures a capacity limit, though solubility is an equilibrium property rather than a fixed mechanical seat count.

Real-world example

Crystals left at the bottom of a jar after careful mixing may indicate that the liquid above is saturated at the current temperature, provided enough time was allowed for equilibrium and no other process interfered. Simply seeing crystals immediately after pouring is not enough; dissolving may still be in progress.

Why?

Why must temperature accompany a solubility value? The equilibrium between dispersed solute and a separate phase can shift with temperature. Using a value measured at one temperature to calculate another can give the wrong dissolved mass.

Common misconception

“Solubility of 40 g per 100 g water means a 40% mass solution.” The 100 g refers to water alone. The saturated solution includes both masses, making 40/140 ≈ 28.6% solute by mass.

Worked example

At a stated temperature, a salt has solubility 24 g per 100 g water. Add 20 g salt to 50 g water and allow equilibrium. Capacity is 24 × 50/100 = 12 g. Thus 12 g dissolves, 8 g remains solid, and the saturated liquid solution has mass 50 + 12 = 62 g. Its mass fraction is 12/62 ≈ 0.194, or 19.4%. Using 20 g in the fraction would wrongly count undissolved solid as part of the liquid.

Quick check

1. Does excess undissolved solid imply that all added solute appears in the solution concentration? Answer: No. Only the dissolved portion contributes to the concentration of the liquid phase; the excess is a separate solid.

Exam focus

Keep three masses separate: solvent, dissolved solute and added solute. Scale solubility with the first, then determine the second from the limit.

Advanced insight

For sparingly soluble salts, an equilibrium constant can describe saturation in terms of dissolved-ion activities. This is a more detailed model than a grams-per-100-grams table and is sensitive to other ions in solution.

Summary

Solubility is a condition-dependent equilibrium capacity. Its numerical basis must be read exactly. Scale it to the available solvent, compare with material added, and exclude excess solid from the mass of the saturated liquid solution.

Practice questions

1. A solute's solubility is 30 g per 100 g water. What is its capacity in 200 g water at the same temperature? Answer: Twice the solvent mass gives twice the limit: 60 g dissolved solute at equilibrium, under the same conditions. 2. Add 25 g to 50 g water when the limit is 30 g per 100 g water. How much remains solid? Answer: The water can hold 15 g, so 25 − 15 = 10 g remains undissolved after equilibrium. 3. What is the saturated liquid's mass fraction in that case? Answer: The liquid contains 15 g solute and 50 g water, so mass fraction is 15/65 ≈ 0.231, or 23.1%.