Preparing Salts by Acid–Base Reaction

Introduction

Making a salt begins with its target ions. An acid can provide the anion and a base the cation; their reaction may form water as a second product. The equation alone is not a complete preparation plan. If the salt remains dissolved, the method must also separate it from water and from any excess acid or base.

Core explanation

To prepare sodium nitrate, choose nitric acid for NO₃⁻ and sodium hydroxide for Na⁺: HNO₃ + NaOH → NaNO₃ + H₂O. The formula follows charge balance, and the equation is one-to-one. Both starting solutions and sodium nitrate are soluble, so the clear mixture cannot be filtered to remove an excess dissolved reactant. A titration can establish the needed acid and alkali volumes. One may then combine measured fresh portions in that ratio, avoiding indicator contamination if pure product is desired, and concentrate the resulting salt solution to crystallise it under suitable conditions.

For an insoluble base, a different approach is possible. To prepare copper(II) sulfate solution, react dilute sulfuric acid with copper(II) oxide: H₂SO₄ + CuO → CuSO₄ + H₂O. Copper(II) oxide does not dissolve appreciably as a simple aqueous hydroxide. Adding a slight excess solid helps consume the acid; unreacted solid can then be filtered away. The filtrate contains dissolved copper(II) sulfate, which can be concentrated and cooled to form crystals. The reaction and separation stages explain why insoluble and soluble reagents call for different preparation routes.

Why not use excess NaOH and filter it? Dissolved NaOH passes through ordinary filter paper along with dissolved sodium nitrate, so filtration cannot separate them. In contrast, an unreacted insoluble oxide or carbonate can remain as a solid and be removed. The physical state of excess reactant determines the separation strategy. A preparation question should therefore name both the chemical reaction and the purification logic.

An acid and base must also be matched to the target salt. Hydrochloric acid supplies chloride; nitric acid supplies nitrate; sulfuric acid can supply sulfate at the stated full-neutralisation endpoint. Sodium hydroxide supplies Na⁺, potassium hydroxide supplies K⁺, and appropriate metal oxides or hydroxides supply other cations. Use ion charges to write the salt formula, then balance water production. For example, 2HCl + CuO → CuCl₂ + H₂O creates copper(II) chloride, not copper(I) chloride, because CuO contains Cu²⁺ in the standard ionic assignment.

Crystallisation must be controlled so a soluble product is recovered without simply drying a mixture of impurities. Evaporation removes some water; cooling may reduce solubility and form crystals. The useful temperature path depends on the particular salt's solubility and stability. Overheating can decompose some compounds or change hydration, so a generic “boil to dryness” instruction is not universally sound. The next pages compare precipitation and crystal recovery in more detail.

Step-by-step reasoning

1. Write the target salt ions and construct a neutral formula. 2. Select acid and base reagents supplying those ions and balance their reaction. 3. Decide whether the base and product are soluble under the stated conditions. 4. If both reactants are soluble, use stoichiometric volumes to avoid dissolved excess; if one is an insoluble solid, an excess can often be filtered away. 5. Recover dissolved product by a suitable crystallisation method and explain impurity control.

Visual explanation

Draw two preparation branches. “Soluble acid + soluble alkali” leads to measured neutralising volumes, then crystallisation. “Acid + excess insoluble oxide” leads to filtration of leftover solid, then crystallisation of dissolved salt. Put a note beside filtration: it removes solid particles, not dissolved excess ions.

Real-world analogy

If two soluble colored liquids are mixed in the wrong amounts, a sieve cannot remove one color from the other. Measuring amounts beforehand prevents the unwanted excess. If one reagent is an undissolved powder, a sieve-like filter can remove what remains. The analogy emphasizes separation state rather than claiming filtration is chemically selective.

Real-world example

In school laboratory planning, a soluble sodium or potassium salt is often prepared using titration to find matching acid and alkali volumes. The recorded equivalence volume is then used for fresh mixtures to reduce indicator dye in the final crystal product. This links quantitative acid–base measurement to practical product purity.

Why?

Why add an insoluble base in excess but avoid excess soluble alkali? Leftover insoluble solid is physically filterable; leftover soluble alkali remains in the filtrate with the desired salt and can contaminate crystals.

Common misconception

“After any acid–base reaction, filter the salt out.” Many salts remain dissolved. Ordinary filtration separates insoluble solids from liquid; it does not collect a dissolved salt until crystallisation or precipitation has produced a solid phase.

Worked example

Plan a conceptual route to potassium chloride from HCl and KOH. The neutral formula is KCl, and HCl + KOH → KCl + H₂O is one-to-one. Because HCl, KOH and KCl can all be in aqueous solution, use measured equivalent amounts rather than excess alkali. Obtain the KCl-containing solution, then remove water under a suitable crystallisation scheme. Filtering the clear reaction mixture before crystallisation would not isolate dissolved KCl.

Quick check

1. Why is titration useful before preparing a soluble salt from a soluble acid and alkali? Answer: It finds matching reactant volumes so neither soluble reagent remains in excess and contaminates the product solution.

Exam focus

Give both a balanced equation and a separation method. State whether the product and any excess reagent are dissolved or solid. Use filtration only for an insoluble solid and crystallisation for recovery of a soluble salt.

Advanced insight

Product purity can be limited by co-crystallising ions, retained mother liquor and changes in crystal hydration. A carefully chosen wash and drying method can improve isolation, but it must not dissolve away much product or alter its chemical form. The reaction equation alone cannot predict these physical details.

Summary

Acid–base salt preparation starts by choosing reagents that supply the desired ions and balancing the reaction. Soluble reagents require controlled amounts; an insoluble excess can often be filtered off. Dissolved salt is generally recovered by a suitable crystallisation step rather than by filtering the clear liquid.

Practice questions

1. Which acid provides nitrate ions for a sodium nitrate preparation? Answer: Nitric acid supplies NO₃⁻; sodium hydroxide or another suitable sodium base supplies Na⁺. 2. Why can excess CuO be removed after reacting with sulfuric acid? Answer: Unreacted copper(II) oxide remains an insoluble solid, so it can be separated by filtration from dissolved copper(II) sulfate. 3. Can ordinary filtration remove excess dissolved KOH from KCl solution? Answer: No. Dissolved KOH ions pass through filter paper with dissolved KCl; stoichiometric control is needed.