Why Metal Reactions with Water Produce Basic Solutions

A small piece of calcium is placed in water. Bubbles appear, the liquid turns cloudy, and a pH probe records an increase in pH. The water has become basic.

But why? Calcium is a metal, hydrogen gas is neutral, and neither seems to explain the hydroxide ions responsible for the change. The answer lies in what happens to water molecules when they gain electrons from the metal.

Before reading on, make a prediction: when a metal reacts with liquid water, which part of the water molecule eventually becomes hydrogen gas, and which part remains in solution?

The hydrogen atoms form \(\ce{H2}\), while hydroxide ions, \(\mathrm{OH}^{-}\), remain in the water. Those hydroxide ions are what make the solution basic.

01Start with the experiment

Consider an investigation designed to find out whether a metal reacting with water actually produces a basic solution.

A small piece of calcium is placed in a beaker containing a measured volume of distilled water. A pH probe sits in the liquid, away from the metal. In a second, matched trial, a small piece of calcium reacts with the same volume of water in a conical flask connected through a delivery tube to a gas syringe.

The first arrangement measures pH. The second measures the volume of gas produced.

What should the experiment show?

If the metal reacts with water through a redox reaction, we expect two important changes:

• Hydrogen gas should form and escape from the liquid.
• Hydroxide ions should enter the aqueous phase, increasing its pH.

The expected observations are:

The pH measurement is particularly important.

Imagine that the original investigation only counted bubbles. It could show that a gas was produced, but it could not establish that the remaining liquid was basic.

The revised investigation adds a pH probe and a water-only control. The control checks that the starting water does not undergo a similar pH change without the metal.

For a fair comparison between different metals, the water temperature, volume, metal mass, exposed surface area, and observation time must also be controlled.

This is an illustrative experimental design, rather than a report of a particular historical experiment. It demonstrates how different measurements answer different chemical questions.

02Where do the hydroxide ions come from?

The key is that a metal reacting with water is a redox reaction.

A redox reaction involves the transfer of electrons. One substance loses electrons, while another gains them.

When a metal reacts, its atoms can lose electrons and become positive ions. This is oxidation.

For a general metal \(M\) forming an ion with charge \(n+\):

\[ \ce{M -> M^{n+} + ne^-} \]

Here:

• \(M\) represents a metal atom.
• \(\mathrm{M}^{n+}\) represents the positive metal ion.
• \(n\) is the number of electrons lost.
• \(\ce{e^-}\) represents an electron.

But where do those electrons go?

Water molecules accept them. During this reduction process, water produces hydrogen gas and hydroxide ions:

\[ \ce{2H2O + 2e^- -> H2 + 2OH^-} \]

This half-equation is the central idea behind the formation of a basic solution.

Notice the two products:

• \(\ce{H2}\) is hydrogen gas, which escapes as bubbles.
• \(\mathrm{OH}^{-}\) remains in the aqueous solution.

The metal is supplying electrons, but the hydroxide ions are produced from the water molecules themselves.

Applying the mechanism to calcium

Calcium loses two electrons to form calcium ions:

\[ \ce{Ca -> Ca^2+ + 2e^-} \]

Water accepts those two electrons:

\[ \ce{2H2O + 2e^- -> H2 + 2OH^-} \]

Combining the half-equations gives:

\[ \ce{Ca + 2H2O -> Ca^2+ + 2OH^- + H2} \]

The calcium ions and hydroxide ions can then form calcium hydroxide:

\[ \ce{Ca^2+ + 2OH^- -> Ca(OH)2} \]

The overall reaction is:

\[ \boxed{\ce{Ca(s) + 2H2O(l) -> Ca(OH)2 + H2(g)}} \]

Calcium hydroxide is a base. Some dissolves in water, releasing hydroxide ions into the aqueous phase.

This increases the hydroxide ion concentration and makes the solution basic.

At \(25^\circ\mathrm{C}\), a basic solution has a pH greater than 7.

The important distinction is that hydrogen gas is not responsible for the increased pH. The dissolved hydroxide ions are.

What happens with sodium?

Sodium follows the same general process, but forms a metal ion with a \(1+\) charge.

Its balanced reaction with water is:

\[ \ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)} \]

Sodium hydroxide is highly soluble in water:

\[ \ce{NaOH(aq) -> Na+(aq) + OH^-(aq)} \]

This means that the hydroxide ions produced are readily available in the solution.

The result is a strongly basic solution.

The reactions of calcium and sodium therefore have the same underlying redox mechanism, even though their metal ions and hydroxide products behave differently.

03Why some metal reactions produce clearer solutions than others

Here is an important complication.

A metal can produce hydroxide ions during its reaction with water, but that does not mean all the hydroxide remains dissolved.

Different metal hydroxides have different solubilities.

Solubility describes how much of a substance can dissolve in a particular solvent under specified conditions.

Consider three familiar hydroxides.

Why cloudy water can still be basic

Consider calcium hydroxide.

It establishes an equilibrium between the undissolved solid and its dissolved ions:

\[ \ce{Ca(OH)2(s) <=> Ca^2+(aq) + 2OH^-(aq)} \]

Some calcium hydroxide dissolves, releasing hydroxide ions.

However, if the solution reaches its solubility limit, additional calcium hydroxide remains as a solid.

This explains why the mixture can look cloudy while the liquid itself is basic.

A pH probe measures the aqueous phase. It does not measure the total amount of hydroxide present in both the dissolved and undissolved forms.

This gives us an important distinction:

The amount of solid metal hydroxide produced is not necessarily the same as the concentration of hydroxide ions in solution.

Think of the undissolved hydroxide as a reservoir that can release ions into the surrounding water. The analogy is useful because the solid can supply more dissolved ions, but it has a limit set by solubility equilibrium. It is not an unlimited supply.

Does a metal always produce a basic solution?

No. The conditions and products matter.

Magnesium provides a useful contrast.

With liquid water, magnesium can form magnesium hydroxide and hydrogen:

\[ \ce{Mg(s) + 2H2O(l) -> Mg(OH)2(s) + H2(g)} \]

However, the reaction is very slow with cold water. Magnesium hydroxide is also sparingly soluble, so relatively little hydroxide enters the liquid.

With steam, the products are different:

\[ \ce{Mg(s) + H2O(g) -> MgO(s) + H2(g)} \]

The product is magnesium oxide, not dissolved magnesium hydroxide.

There is no basic aqueous solution being produced in the hot reaction zone because the reaction involves steam and produces a solid oxide.

If magnesium oxide is subsequently placed in liquid water, it can react to form magnesium hydroxide:

\[ \ce{MgO(s) + H2O(l) -> Mg(OH)2(s)} \]

Some of that hydroxide can then dissolve and make the water basic.

The distinction between liquid water and steam is therefore important. A metal producing an oxide and hydrogen gas does not automatically mean that a basic solution has formed.

04How the activity series helps explain the reaction

The activity series ranks metals according to their tendency to undergo oxidation.

A metal higher in the series is generally more readily oxidised and acts as a stronger reducing agent.

This helps predict whether a metal is likely to participate in a redox reaction.

However, the activity series does not tell you exactly how quickly a particular metal will react with cold water.

That is a separate question.

Consider three different measurements:

These measurements are related, but they are not interchangeable.

For example, a metal might have a strong tendency to oxidise but show almost no visible reaction because its surface is protected by an oxide layer.

The effect of surface coatings

Aluminium is a useful example.

Although aluminium is relatively high in the activity series, it often appears unreactive when placed in cold water.

Why?

Aluminium naturally develops a thin layer of aluminium oxide, \(\ce{Al2O3}\), on its surface.

This layer separates the underlying metal from the surrounding water, restricting the contact needed for the reaction to proceed.

The phenomenon is called passivation.

This introduces an important distinction between thermodynamics and kinetics.

• Thermodynamics considers whether a reaction is energetically favourable under specified conditions.
• Kinetics considers how quickly the reaction actually happens.

A reaction can be thermodynamically favourable but proceed extremely slowly because of a large activation energy or a protective surface layer.

Surface area also matters. A metal with a larger exposed surface provides more sites where water molecules can interact with the metal.

Increasing exposed surface area can increase the reaction rate without changing the metal’s position in the activity series.

For a more detailed comparison of these ideas, see How to Compare Metal Reactivity Without Confusing Reaction Rate.

The broader patterns of cold-water reactions, steam reactions, and metal activity are covered in Metal Reactivity with Water: HSC Chemistry Guide.

07A useful decision process for HSC questions

When you are asked why a metal reacting with water produces a basic solution, work through these four decisions.

1. Identify the reaction conditions. Is the metal reacting with cold water, hot liquid water, or steam?

2. Identify the redox products. Is water being reduced to hydrogen gas and hydroxide ions?

3. Consider hydroxide solubility. Does the metal hydroxide dissolve readily, partially, or very poorly?

4. Consider kinetic limitations. Could an oxide coating, hydroxide deposit, or small exposed surface area explain a slow or invisible reaction?

This prevents you from making the common mistake of assuming that every metal that reacts with water must produce the same type of basic solution.

08Questions and solutions

Question 1

A piece of calcium is added to cold water. Hydrogen bubbles form, and universal indicator changes from green towards blue.

Write a balanced chemical equation for the reaction and explain why the solution becomes basic.

Solution 1

The solution becomes basic because hydroxide ions are produced when water accepts electrons from calcium.

The balanced equation is:

\[ \ce{Ca(s) + 2H2O(l) -> Ca(OH)2 + H2(g)} \]

Calcium loses two electrons to form \(\mathrm{Ca}^{2+}\), while water gains electrons and forms hydrogen gas and hydroxide ions.

The hydroxide ions remain partly dissolved in the water, increasing its pH.

The indicator change provides evidence that the liquid has become basic. The hydrogen bubbles are evidence of gas production, not the cause of the increased pH.

Question 2

A student reacts calcium with water and obtains a cloudy mixture. They filter the mixture, collecting \(250.0\ \mathrm{mL}\) of clear filtrate.

The filtrate has a pH of 11.00 at \(25^\circ\mathrm{C}\).

Calculate:

(a) The hydroxide ion concentration in the filtrate.

(b) The amount of hydroxide ions present in the filtrate.

(c) The amount of calcium hydroxide dissolved in the filtrate.

Explain why the filtrate can remain basic even though a white solid was removed during filtration.

Solution 2

The filtrate is still basic because some calcium hydroxide dissolved before filtration, releasing hydroxide ions into the water.

Step 1: Calculate pOH.

At \(25^\circ\mathrm{C}\):

\[ \begin{aligned} \mathrm{pOH}&=14.00-\mathrm{pH}\\ &=14.00-11.00\\ &=3.00 \end{aligned} \]

Step 2: Calculate hydroxide ion concentration.

\[ [\mathrm{OH}^{-}]=10^{-\mathrm{pOH}} \]

\[ [\mathrm{OH}^{-}]=10^{-3.00} \]

\[ \boxed{[\mathrm{OH}^{-}]=1.00\times10^{-3}\ \mathrm{mol\,L^{-1}}} \]

Step 3: Calculate the amount of hydroxide ions.

Convert the volume:

\[ 250.0\ \mathrm{mL}=0.2500\ \mathrm{L} \]

Using \(n=cV\):

\[ \begin{aligned} n(\mathrm{OH}^{-})&=(1.00\times10^{-3})(0.2500)\\ &=2.50\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the amount of dissolved calcium hydroxide.

The dissolution equation is:

\[ \ce{Ca(OH)2(s) <=> Ca^2+(aq) + 2OH^-(aq)} \]

One mole of calcium hydroxide produces two moles of hydroxide ions.

Therefore:

\[ \begin{aligned} n(\ce{Ca(OH)2})&=\frac{2.50\times10^{-4}}{2}\\ &=1.25\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Step 5: Explain the remaining solid.

Calcium hydroxide is only slightly soluble in water. Some dissolves, producing calcium ions and hydroxide ions, while the remainder stays as a solid.

Filtration removes the undissolved solid but does not remove the ions already dissolved in the filtrate.

The filtrate therefore remains basic.

The calculated amount represents the dissolved calcium hydroxide, not necessarily the total amount of calcium hydroxide produced during the original reaction.

Question 3

Magnesium is heated in a tube while steam passes over its surface.

Hydrogen gas is collected at the outlet, and a white solid remains in the heated section.

A student argues that the collected products must include a basic solution because magnesium is above hydrogen in the activity series.

Explain why this conclusion is incorrect.

Write the balanced chemical equation and describe what would happen if the white solid were subsequently placed in liquid water.

Solution 3

The reaction in the heated tube does not directly produce a basic aqueous solution because its products are solid magnesium oxide and hydrogen gas, not dissolved magnesium hydroxide.

The balanced equation is:

\[ \ce{Mg(s) + H2O(g) -> MgO(s) + H2(g)} \]

The magnesium is oxidised, while water is reduced to hydrogen gas.

However, the oxygen-containing product is magnesium oxide, which remains solid under the reaction conditions.

There is no liquid solution in the hot reaction zone receiving dissolved hydroxide ions.

If the magnesium oxide is subsequently added to liquid water, it can react to form magnesium hydroxide:

\[ \ce{MgO(s) + H2O(l) -> Mg(OH)2(s)} \]

Some magnesium hydroxide can dissolve, releasing hydroxide ions and making the aqueous phase basic.

The important distinction is between producing a solid metal oxide using steam and producing dissolved hydroxide ions in liquid water.

Being high in the activity series does not mean every reaction with water produces the same products.

Question 4

Two magnesium strips have identical masses and exposed geometric surface areas. They are placed in separate containers containing equal volumes of water at \(60^\circ\mathrm{C}\).

One strip has a freshly exposed surface. The other has a pre-existing magnesium hydroxide coating.

The following illustrative data are obtained:

Both strips still contain unreacted metal after 10 minutes.

(a) Which strip has the greater observed reaction rate?

(b) Explain why the two strips behave differently despite being the same metal.

(c) Predict what might happen to the reaction rate if the coating on the second strip develops cracks.

(d) Explain why these results do not establish that the freshly exposed magnesium has a different position in the activity series.

Solution 4

The freshly exposed magnesium has the greater observed reaction rate because it produces substantially more hydrogen over the same time interval.

Step 1: Compare hydrogen production.

After two minutes:

• Fresh surface: \(2.4\ \mathrm{mL}\).
• Coated surface: \(0.2\ \mathrm{mL}\).

The freshly exposed strip produces twelve times the measured gas volume during this interval.

Because both samples have the same mass, metal identity, water volume, temperature, and exposed geometric area, the difference points towards their surface conditions.

Step 2: Explain the effect of the coating.

The magnesium hydroxide coating creates a barrier between the underlying magnesium and water.

This reduces the accessibility of the metal surface and slows the reaction.

The freshly exposed sample allows water molecules to interact more readily with the magnesium surface.

Step 3: Predict the effect of cracking.

If the coating develops cracks, water can reach previously covered regions of magnesium.

This increases the accessible surface area and may increase the reaction rate.

However, additional magnesium hydroxide can form as the reaction proceeds, potentially slowing the reaction again.

Step 4: Distinguish surface condition from intrinsic reactivity.

Both strips are magnesium. Their position in the activity series is identical.

The difference in gas production is a kinetic effect caused by the surface coating, not a change in the metal’s intrinsic tendency to undergo oxidation.

This is why reaction rates must be compared using controlled surface conditions.

Question 5

A student wants to compare the basicity of solutions produced when sodium and magnesium react with water.

For a theoretical stoichiometric comparison, assume that separate samples containing \(0.0100\ \mathrm{mol}\) of each metal react completely with excess liquid water.

The final aqueous volume in each case is \(0.100\ \mathrm{L}\), and the temperature is \(25^\circ\mathrm{C}\).

Use the following information:

• Sodium hydroxide dissolves completely.
• Magnesium hydroxide is sparingly soluble.
• In the magnesium trial, excess solid \(\ce{Mg(OH)2}\) remains, and the saturated solution has a hydroxide ion concentration of \(2.0\times10^{-4}\ \mathrm{mol\,L^{-1}}\).
• Assume no carbon dioxide enters either solution and treat both solutions as ideal.

In a separate time-course observation, the magnesium solution reaches pH 10.3 and then maintains approximately this pH, even though hydrogen gas continues to be produced.

Answer the following:

(a) Calculate the pH of the sodium hydroxide solution.

(b) Calculate the pH of the saturated magnesium hydroxide solution.

(c) Calculate the theoretical amount of hydrogen produced by each metal.

(d) A student concludes that magnesium must be more reactive because it produces twice as much hydrogen per mole of metal. Evaluate this claim using the activity series and the balanced equations.

(e) Explain how the magnesium reaction can continue producing hydrogen while the pH remains approximately constant.

This is a theoretical calculation, not a practical experiment. Sodium reacts vigorously with water and must not be used for an ordinary student water-reaction test.

Solution 5

Sodium produces the higher-pH solution, while magnesium produces twice as much hydrogen per mole of metal. Neither result demonstrates that magnesium is more reactive. Sodium has the greater tendency to undergo oxidation according to the activity series.

Step 1: Calculate the pH of the sodium hydroxide solution.

The balanced reaction is:

\[ \ce{2Na + 2H2O -> 2NaOH + H2} \]

The mole ratio between sodium and sodium hydroxide is \(1:1\).

Therefore:

\[ n(\ce{NaOH})=0.0100\ \mathrm{mol} \]

Because sodium hydroxide dissolves completely:

\[ [\mathrm{OH}^{-}]=\frac{n}{V} \]

\[ [\mathrm{OH}^{-}]=\frac{0.0100}{0.100} \]

\[ [\mathrm{OH}^{-}]=0.100\ \mathrm{mol\,L^{-1}} \]

Calculate pOH:

\[ \mathrm{pOH}=-\log_{10}(0.100)=1.00 \]

Therefore:

\[ \mathrm{pH}=14.00-1.00 \]

\[ \boxed{\mathrm{pH}=13.00} \]

Step 2: Calculate the pH of the magnesium hydroxide solution.

The saturated hydroxide concentration is provided:

\[ [\mathrm{OH}^{-}]=2.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]

Calculate pOH:

\[ \begin{aligned} \mathrm{pOH}&=-\log_{10}(2.0\times10^{-4})\\ &=3.70 \end{aligned} \]

Therefore:

\[ \mathrm{pH}=14.00-3.70 \]

\[ \boxed{\mathrm{pH}=10.30} \]

The magnesium hydroxide solution has a lower pH despite the fact that each mole of magnesium can produce twice as many hydroxide ions as each mole of sodium.

The reason is solubility.

Most of the magnesium hydroxide remains as a solid rather than dissolving into the water.

The pH depends on the dissolved hydroxide concentration, not the total amount of hydroxide produced.

Step 3: Calculate the amount of hydrogen produced.

For sodium:

\[ \ce{2Na + 2H2O -> 2NaOH + H2} \]

Two moles of sodium produce one mole of hydrogen.

\[ n(\ce{H2})=\frac{0.0100}{2} \]

\[ \boxed{n(\ce{H2})=0.00500\ \mathrm{mol}} \]

For magnesium:

\[ \ce{Mg + 2H2O -> Mg(OH)2 + H2} \]

One mole of magnesium produces one mole of hydrogen.

\[ \boxed{n(\ce{H2})=0.0100\ \mathrm{mol}} \]

Therefore, magnesium produces twice as much hydrogen for the same initial amount of metal.

Step 4: Evaluate the student’s claim about reactivity.

The student’s conclusion is incorrect.

The amount of hydrogen produced depends on the balanced equation and the amount of metal that reacts.

Sodium loses one electron per atom, while magnesium loses two electrons per atom.

Consequently, one mole of magnesium produces twice as much hydrogen as one mole of sodium.

This is a stoichiometric difference, not evidence that magnesium is more reactive.

The activity series places sodium above magnesium. Sodium is more readily oxidised and has a greater tendency to act as a reducing agent.

The total amount of hydrogen produced also does not establish which reaction has the greater rate.

To compare reaction rates, we would need to measure hydrogen production per unit time under appropriately controlled conditions.

Step 5: Explain the constant pH during the magnesium reaction.

The magnesium solution becomes saturated with magnesium hydroxide.

Once saturation is reached, additional magnesium hydroxide produced by the reaction largely remains as a solid rather than increasing the dissolved hydroxide concentration.

The dissolution equilibrium is:

\[ \ce{Mg(OH)2(s) <=> Mg^2+(aq) + 2OH^-(aq)} \]

As long as excess solid remains and the temperature is constant, the dissolved hydroxide concentration stays approximately at its saturation value.

This means the pH can remain around 10.3 even while more magnesium reacts.

Hydrogen production can continue because the metal is still being oxidised and water is still being reduced.

The reaction may become slower as the magnesium hydroxide coating builds up, but a constant pH does not prove that the reaction has stopped.

This final example brings together three different ideas:

• The activity series describes the tendency of a metal to undergo oxidation.
• Reaction rate describes how quickly hydrogen is produced.
• Solubility determines how much hydroxide remains dissolved and therefore influences pH.

Confusing these measurements leads to incorrect conclusions about metal reactivity.

The next useful step is to connect hydroxide concentration with solubility equilibrium and solution concentration calculations. That allows you to predict not just whether a reaction produces a basic solution, but how basic the solution can actually become.