Acids Reacting with Metal Hydroxides

Introduction

Sodium hydroxide dissolves readily in water to give a strongly alkaline solution. Copper(II) hydroxide is a pale blue solid that hardly dissolves at all. Yet both neutralise acids in exactly the same way, producing a salt and water. Metal hydroxides are the most direct bases of all, because they carry ready-made hydroxide ions. This page compares how soluble and insoluble hydroxides react with acids.

Core explanation

The general equation.

metal hydroxide + acid → salt + water

This is the same pattern as for metal oxides, but here the base provides hydroxide ions, OH⁻, which combine with H⁺ one-to-one.

Soluble hydroxides (alkalis). Group 1 hydroxides such as sodium, potassium and lithium hydroxide dissolve well. Calcium hydroxide dissolves slightly. Their solutions are alkalis, and the reaction takes place between two solutions:

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)

Ca(OH)₂(aq) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + 2H₂O(l)

The ionic equation is simply H⁺(aq) + OH⁻(aq) → H₂O(l). There is no visible change — both solutions are colourless and so is the product — so an indicator or pH meter is needed to show when neutralisation is complete. Because an alkali and an acid can be measured precisely by volume, these reactions are the basis of titration.

Insoluble hydroxides. Hydroxides of most transition metals, such as copper(II), iron(II) and iron(III), and also magnesium hydroxide, are insoluble. They still react with acids:

Cu(OH)₂(s) + 2HCl(aq) → CuCl₂(aq) + 2H₂O(l)

Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l)

Fe(OH)₃(s) + 3HNO₃(aq) → Fe(NO₃)₃(aq) + 3H₂O(l)

Here the solid visibly disappears as it reacts, and a coloured salt solution may form (blue for copper). The ionic equation includes the solid, for example:

Cu(OH)₂(s) + 2H⁺(aq) → Cu²⁺(aq) + 2H₂O(l)

Comparing the two types.

Feature Soluble hydroxide Insoluble hydroxide --- --- --- Example NaOH Cu(OH)₂ State in equation (aq) (s) Visible change None without indicator Solid disappears How to know it is complete Indicator changes colour Excess solid remains Typical use in salt making Titration Add excess, then filter

Counting water. Each OH⁻ ion reacts with one H⁺ ion to make one water molecule. Ca(OH)₂ has two hydroxide ions, so it forms two water molecules and needs two H⁺ ions.

Step-by-step reasoning

To write a hydroxide–acid equation:

1. Find the metal ion charge from the number of OH groups (NaOH → Na⁺; Fe(OH)₃ → Fe³⁺). 2. Combine the metal ion with the acid's anion to make the salt. 3. Write hydroxide + acid → salt + water. 4. Make the number of H⁺ equal the number of OH⁻; that number is also the number of H₂O. 5. Use (aq) for soluble hydroxides and (s) for insoluble ones.

Visual explanation

In the neutralisation simulation, compare two beakers. In the first, OH⁻ dots from dissolved NaOH meet H⁺ dots throughout the liquid, and only the indicator colour shows what is happening. In the second, H⁺ dots attack the surface of a blue Cu(OH)₂ solid, which shrinks as its ions are carried into solution.

Real-world analogy

A soluble hydroxide is like a crowd of volunteers already spread through a town, able to help anywhere at once. An insoluble hydroxide is like volunteers waiting at a single base: they still help, but only those at the doorway can get to work at any moment.

Real-world example

"Milk of magnesia", a suspension of magnesium hydroxide, is taken for indigestion. Because it is almost insoluble, it does not make the stomach strongly alkaline; it neutralises excess hydrochloric acid gradually, forming magnesium chloride and water.

Why?

Why is an indicator needed with sodium hydroxide but not with copper(II) hydroxide? Sodium hydroxide, hydrochloric acid and sodium chloride solutions all look the same, so there is nothing to see. With copper(II) hydroxide, the disappearing solid and appearing blue colour show the reaction directly.

Common misconception

"Insoluble hydroxides cannot react because they do not produce OH⁻ ions in water." The OH⁻ ions are locked in the solid, but acid attacks the solid's surface directly. The hydroxide ions react there and the solid gradually dissolves as a salt.

Worked example

Question: Write a balanced equation with state symbols for iron(II) hydroxide reacting with sulfuric acid.

Reasoning: Fe(OH)₂ contains Fe²⁺; with SO₄²⁻ the salt is FeSO₄. Two OH⁻ need two H⁺, which one H₂SO₄ supplies, forming two H₂O. Iron(II) hydroxide is insoluble, so it is (s).

Answer: Fe(OH)₂(s) + H₂SO₄(aq) → FeSO₄(aq) + 2H₂O(l).

Quick check

1. How many water molecules form when one Ca(OH)₂ unit is neutralised? Answer: Two, because it contains two hydroxide ions.

Exam focus

Know the general equation and be able to choose the right state symbol for the hydroxide. Explain why an indicator is needed for soluble hydroxides. Link soluble hydroxides to titration and insoluble ones to the "excess and filter" method of salt preparation.

Advanced insight

Some insoluble hydroxides, such as zinc and aluminium hydroxide, are amphoteric: they dissolve in excess sodium hydroxide as well as in acids. This behaviour is used in qualitative analysis to tell these metal ions apart from others that form similar white precipitates.

Summary

Metal hydroxides react with acids to give a salt and water, with each OH⁻ neutralising one H⁺. Soluble hydroxides (alkalis) react in solution with no visible change, so an indicator is used, making them suitable for titration. Insoluble hydroxides react as solids that visibly disappear, so they are added in excess and filtered.

Practice questions

1. Complete the word equation: potassium hydroxide + nitric acid → ? Answer: Potassium nitrate + water. 2. Write a balanced equation for magnesium hydroxide reacting with hydrochloric acid. Answer: Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l). 3. Explain why an indicator is used when sodium hydroxide solution neutralises hydrochloric acid. Answer: All the solutions involved are colourless, so there is no visible sign of when neutralisation is complete; an indicator changes colour at that point. 4. Write the ionic equation for copper(II) hydroxide reacting with an acid. Answer: Cu(OH)₂(s) + 2H⁺(aq) → Cu²⁺(aq) + 2H₂O(l).