Preparing a Soluble Salt from an Insoluble Base

Introduction

Many useful salts, such as copper(II) sulfate, zinc sulfate and magnesium chloride, dissolve in water. To make a pure, dry sample of one of them we need a method that leaves nothing else in the final solution. A neat trick is to react an acid with an insoluble base, such as a metal oxide or a metal carbonate. Because the base does not dissolve, any unreacted base can simply be filtered off, leaving a solution that contains only the salt and water.

Core explanation

The general reaction. An acid reacts with a base to give a salt and water. With a metal carbonate, carbon dioxide is also produced:

- acid + metal oxide → salt + water - acid + metal carbonate → salt + water + carbon dioxide

For example, copper(II) oxide reacts with dilute sulfuric acid to give copper(II) sulfate:

CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)

Why use excess base? The goal is a solution containing only the salt. If we stopped before all the acid had reacted, the solution would still contain leftover acid, and the crystals would be contaminated with it. So the insoluble base is added a little at a time, with stirring, until no more will react and some solid remains visible. At that point all the acid has been used up. The leftover solid is the excess — and because it is insoluble, it is easy to remove.

Signs that the acid is used up. With a metal carbonate, fizzing stops when no acid remains. With a metal oxide, unreacted black or white powder stays in the beaker however much you stir. Warming the acid gently speeds up the reaction with oxides, which can be slow at room temperature.

Filtration. The mixture is filtered. The unreacted base stays on the filter paper as the residue ; the salt solution passes through as the filtrate . The filtrate now contains just the salt dissolved in water.

Crystallisation. The filtrate is warmed to evaporate some of the water until the solution is saturated — a crystal will form on a cold glass rod dipped in and removed. The solution is then left to cool slowly. As it cools, the salt becomes less soluble and crystals grow. Slow cooling gives larger, better-shaped crystals. The crystals are separated, rinsed with a little cold water and dried gently, for example between filter papers.

Why not evaporate to dryness? Heating strongly until all the water has gone can decompose some salts, spit hot solid and destroy the water of crystallisation in hydrated salts. Crystallisation gives purer, well-formed crystals because soluble impurities stay behind in the remaining solution.

When is this method suitable? It works when the salt is soluble and the base is insoluble . It is not used for metals such as sodium or potassium, whose oxides, hydroxides and carbonates are soluble — excess could not be filtered off, so a titration is used instead.

Step-by-step reasoning

To plan the preparation of a named soluble salt:

1. Identify the metal in the salt: it comes from the base. 2. Identify the non-metal part: sulfate comes from sulfuric acid, chloride from hydrochloric acid, nitrate from nitric acid. 3. Choose an insoluble base of that metal (oxide, hydroxide or carbonate). 4. React with excess base, filter off the leftover solid, then crystallise the filtrate.

Visual explanation

Picture three stages side by side. First, a beaker of colourless acid with black powder at the bottom that stays even after stirring. Next, a funnel and filter paper: black solid trapped on the paper, a clear blue liquid dripping into a basin. Finally, the basin after cooling, with blue diamond-shaped crystals sitting in a little leftover solution.

Real-world analogy

Making the salt is like sweetening tea with sugar lumps that do not dissolve completely. You keep adding lumps until one stays whole, so you know the tea cannot take any more — then you fish out the spare lump, leaving exactly-saturated tea behind.

Real-world example

Zinc sulfate, used in dietary supplements and to treat zinc deficiency in crops, can be made by reacting zinc oxide or zinc carbonate with sulfuric acid. Industrial processes use the same logic: excess solid base ensures no acid is left in the product, and filtering removes the unreacted solid.

Why?

Why must the base be insoluble for this method? The whole point of adding excess is that the leftover reactant can be removed by filtration. A soluble base would dissolve in the solution, so its excess could not be filtered out and would contaminate the salt.

Common misconception

"The crystals form because the water is boiled away completely." In good practice, only part of the water is evaporated. Crystals form as the saturated solution cools, because the salt is less soluble at lower temperatures.

Worked example

Question: Outline how to make crystals of magnesium chloride starting from magnesium carbonate.

Reasoning: Magnesium comes from the carbonate; chloride comes from hydrochloric acid. MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) + CO₂(g). Add the insoluble carbonate to the acid until fizzing stops and solid remains; filter; warm the filtrate to concentrate it; cool to crystallise.

Answer: React excess magnesium carbonate with dilute hydrochloric acid, filter off the excess carbonate, then concentrate and cool the filtrate to form magnesium chloride crystals.

Quick check

1. Why is excess copper(II) oxide added when making copper(II) sulfate from sulfuric acid? Answer: To make sure all the acid reacts, so the final solution contains only copper(II) sulfate and water.

Exam focus

Examiners expect three named stages: add excess insoluble base (and say how you know it is in excess), filter to remove the excess, then crystallise the filtrate. Always give the balanced equation and state the colour changes where relevant.

Advanced insight

The yield of crystals is never 100%. Some salt remains dissolved in the cold solution left after crystallisation, called the mother liquor, and some is lost on apparatus. Chemists often concentrate the mother liquor again to collect a second crop of crystals, though that crop is usually less pure than the first.

Summary

A soluble salt can be made by reacting an acid with an excess of an insoluble metal oxide, hydroxide or carbonate. The excess ensures all the acid is used. Filtration removes the unreacted base, and the filtrate is concentrated and cooled so that pure crystals of the salt form.

Practice questions

1. Name the acid and an insoluble base you would use to make zinc nitrate. Answer: Dilute nitric acid with zinc oxide (or zinc carbonate). 2. How can you tell that excess copper(II) carbonate has been added to an acid? Answer: Fizzing stops and some green solid remains undissolved in the beaker. 3. Why is the filtrate not heated until it is completely dry? Answer: Strong heating can decompose the salt or remove its water of crystallisation; cooling a saturated solution gives purer, well-formed crystals. 4. Explain why this method is not used to make sodium sulfate. Answer: Sodium oxide, hydroxide and carbonate are all soluble, so excess base could not be removed by filtration; titration is used instead. 5. Write the equation for copper(II) oxide reacting with hydrochloric acid. Answer: CuO(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l)