How to Calculate Ion Concentration from a Salt Solution

A solution bottle is labelled \(0.200\ \mathrm{mol\,L^{-1}}\) calcium chloride. Does that mean the concentration of chloride ions is also \(0.200\ \mathrm{mol\,L^{-1}}\), or is it \(0.400\ \mathrm{mol\,L^{-1}}\)?

It's \(0.400\ \mathrm{mol\,L^{-1}}\). Each formula unit of calcium chloride produces one calcium ion and two chloride ions when it dissolves:

\[
\ce{CaCl2(aq) -> Ca^{2+}(aq) + 2Cl^{-}(aq)}
\]

Therefore:

\[
[\ce{Ca^{2+}}]=0.200\ \mathrm{mol\,L^{-1}}
\]

\[
[\ce{Cl^-}]=2(0.200)=0.400\ \mathrm{mol\,L^{-1}}
\]

That's the central rule: **to calculate the concentration of an ion from the concentration of a soluble salt, multiply the salt's amount concentration by the stoichiometric coefficient of that ion in the dissociation equation.**

The calculation is straightforward once you know what the salt concentration actually counts. The important part is recognising when you need to convert the given concentration first.

## The basic method

For a soluble ionic salt that dissociates completely, follow these steps:

1. Write the balanced dissociation equation.
2. Identify the coefficient of the ion you are interested in.
3. Multiply the salt concentration by that coefficient.
4. Include the correct units and state any assumptions.

For a general salt that dissociates according to

\[
\ce{A_{x}B_{y} -> xA + yB}
\]

the ion concentrations are:

\[
[\ce{A}]=x\,c_{\text{salt}}
\]

\[
[\ce{B}]=y\,c_{\text{salt}}
\]

Here, \(c_{\text{salt}}\) is the amount concentration of the dissolved salt, measured in \(\mathrm{mol\,L^{-1}}\), while \(x\) and \(y\) are the coefficients from the dissociation equation.

This assumes the salt is sufficiently soluble, dissolves completely, and does not undergo another reaction that removes either ion from solution.

[IMAGE: An annotated particle-level comparison of dissolved calcium chloride and aluminium sulfate. Show one CaCl2 formula unit producing one Ca2+ ion and two Cl- ions, and one Al2(SO4)3 formula unit producing two Al3+ ions and three SO4^2- ions. Use separate, clearly distinguishable hydrated ions dispersed in water, with the 1:2 and 2:3 particle ratios labelled. Distinguish subscripts in the salt formula from superscript ionic charges.]

## What does the salt concentration actually measure?

Before multiplying anything, check how the concentration has been given.

In HSC Chemistry, the concentration of a salt solution will commonly be an **amount concentration**, also called molar concentration. It tells you how many moles of salt formula units are present in each litre of solution.

\[
c=\frac{n}{V}
\]

where:

- \(c\) is amount concentration in \(\mathrm{mol\,L^{-1}}\);
- \(n\) is the amount of dissolved salt in moles;
- \(V\) is the final volume of solution in litres.

Notice that this measures the amount of salt, not the amount of each individual ion. The dissociation equation provides the conversion between them.

However, a question might give you a mass concentration instead.

| Type of concentration | What it tells you | Typical units |
|---|---|---|
| Amount concentration | Moles of salt per litre of solution | \(\mathrm{mol\,L^{-1}}\) |
| Mass concentration | Grams of salt per litre of solution | \(\mathrm{g\,L^{-1}}\) |
| Mass percentage | Grams of salt per 100 g of solution | \(\%\ \mathrm{m/m}\) |
| Volume percentage | Volume of solute per 100 units of solution volume | \(\%\ \mathrm{v/v}\) |

Volume percentage is generally used for mixtures of liquids, rather than a solid ionic salt dissolved in water. A volume by itself is not a concentration.

If the question gives mass concentration, convert it into amount concentration before calculating the ions:

\[
c_{\text{salt}}=
\frac{\text{mass concentration}}{M}
\]

Here, \(M\) is the molar mass of the salt in \(\mathrm{g\,mol^{-1}}\).

For example, a solution containing \(11.1\ \mathrm{g\,L^{-1}}\) of anhydrous calcium chloride has:

\[
c_{\ce{CaCl2}}=
\frac{11.1}{111.0}
=0.100\ \mathrm{mol\,L^{-1}}
\]

Therefore, its chloride-ion concentration is \(0.200\ \mathrm{mol\,L^{-1}}\).

If the concentration is given as a mass percentage, you need to be more careful. A mass percentage describes the mass of salt relative to the total mass of solution, not its volume. To convert it into \(\mathrm{mol\,L^{-1}}\), you need the solution's density.

Do not automatically assume the density is \(1.00\ \mathrm{g\,mL^{-1}}\). That is a useful approximation for water in some situations, but a dissolved salt changes the density of the solution. Use a supplied density or an explicitly stated approximation.

## Worked example: Finding ion concentrations from a molar concentration

A student prepares a \(0.0800\ \mathrm{mol\,L^{-1}}\) solution of aluminium sulfate, \(\ce{Al2(SO4)3}\).

Calculate the concentrations of aluminium ions and sulfate ions, assuming complete dissociation.

**Step 1**: Write the balanced dissociation equation.

\[
\ce{Al2(SO4)3(aq) -> 2Al^{3+}(aq) + 3SO4^{2-}(aq)}
\]

One formula unit produces two aluminium ions and three sulfate ions.

**Step 2**: Find the coefficient of each ion.

- Aluminium ions have a coefficient of 2.
- Sulfate ions have a coefficient of 3.

Remember that the subscript 2 outside the brackets in \(\ce{Al2(SO4)3}\) belongs to aluminium, while the subscript 3 outside the brackets multiplies the entire sulfate group.

**Step 3**: Calculate the aluminium-ion concentration.

\[
[\ce{Al^{3+}}]
=2(0.0800)
=0.160\ \mathrm{mol\,L^{-1}}
\]

**Step 4**: Calculate the sulfate-ion concentration.

\[
[\ce{SO4^{2-}}]
=3(0.0800)
=0.240\ \mathrm{mol\,L^{-1}}
\]

The final concentrations are:

| Ion | Concentration |
|---|---:|
| \(\ce{Al^{3+}}\) | \(0.160\ \mathrm{mol\,L^{-1}}\) |
| \(\ce{SO4^{2-}}\) | \(0.240\ \mathrm{mol\,L^{-1}}\) |

You can check the result using electrical neutrality. The total positive charge concentration is \(3(0.160)=0.480\), while the total negative charge concentration is \(2(0.240)=0.480\), in \(\mathrm{mol}\) of charge per litre.

The charges are balanced, which is a useful way to catch a missing coefficient or an incorrect ion concentration.

## How the concentration is established in the lab

The calculation becomes easier to understand when you picture how a salt solution is actually prepared.

Imagine you need \(250.0\ \mathrm{mL}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) magnesium chloride solution.

You have solid magnesium chloride, distilled water, a balance, a beaker, a funnel, and a \(250.0\ \mathrm{mL}\) volumetric flask.

Each piece of equipment has a different job:

1. The balance measures the mass of magnesium chloride.
2. The beaker holds the salt while it dissolves in a smaller volume of water.
3. The funnel transfers the solution into the volumetric flask.
4. The volumetric flask establishes the final solution volume.
5. The flask is stoppered and inverted several times to mix the solution thoroughly.

[IMAGE: A chronological laboratory apparatus illustration showing a balance weighing solid magnesium chloride in a weighing boat, a beaker containing the dissolved salt, a glass funnel transferring the solution into a 250.0 mL volumetric flask, and the completed flask filled exactly to its calibration mark. Include rinsing of the beaker and funnel to ensure quantitative transfer. Label the balance measurement as mass and the flask calibration mark as final solution volume.]

There is an important detail here.

Suppose you measure exactly \(250.0\ \mathrm{mL}\) of water and then add the salt. Have you made \(250.0\ \mathrm{mL}\) of solution?

Not necessarily. The dissolved salt contributes to the final solution volume, and you cannot assume that the volumes of water and dissolved solute simply add together.

The correct method is to dissolve the salt in less than the required final volume, transfer it to the volumetric flask, and then add water until the bottom of the meniscus reaches the calibration mark.

The first approach does not control the final volume. The revised method does.

This is also why you do not need the solution's density when you prepare a solution using a calibrated volumetric flask. The flask directly establishes the final volume.

One more distinction: this preparation does not directly measure the number of ions. The balance and volumetric flask establish the salt concentration. The dissociation equation then allows us to predict the concentrations of the ions.

### Worked example: Calculate the mass of salt needed to prepare a solution

A student needs to prepare \(250.0\ \mathrm{mL}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) magnesium chloride solution using anhydrous \(\ce{MgCl2}\).

Calculate:

a) The mass of magnesium chloride required.

b) The resulting concentrations of magnesium ions and chloride ions.

**Step 1**: Calculate the amount of magnesium chloride required.

Use:

\[
n=cV
\]

The concentration is \(0.100\ \mathrm{mol\,L^{-1}}\), and the final volume is \(0.2500\ \mathrm{L}\).

\[
\begin{aligned}
n&=0.100\times0.2500\\
&=0.0250\ \mathrm{mol}
\end{aligned}
\]

**Step 2**: Calculate the molar mass of magnesium chloride.

Using the periodic table:

\[
\begin{aligned}
M(\ce{MgCl2})
&=24.31+2(35.45)\\
&=95.21\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

**Step 3**: Convert the required amount into mass.

Use:

\[
m=nM
\]

\[
\begin{aligned}
m&=0.0250\times95.21\\
&=2.38\ \mathrm{g}
\end{aligned}
\]

The student needs \(2.38\ \mathrm{g}\) of anhydrous magnesium chloride.

**Step 4**: Write the dissociation equation.

\[
\ce{MgCl2(aq) -> Mg^{2+}(aq) + 2Cl^{-}(aq)}
\]

One mole of magnesium chloride produces one mole of magnesium ions and two moles of chloride ions.

**Step 5**: Calculate the ion concentrations.

For magnesium ions:

\[
[\ce{Mg^{2+}}]=1(0.100)
=0.100\ \mathrm{mol\,L^{-1}}
\]

For chloride ions:

\[
[\ce{Cl^-}]=2(0.100)
=0.200\ \mathrm{mol\,L^{-1}}
\]

The final answers are:

| Quantity | Result |
|---|---:|
| Mass of \(\ce{MgCl2}\) | \(2.38\ \mathrm{g}\) |
| \([\ce{Mg^{2+}}]\) | \(0.100\ \mathrm{mol\,L^{-1}}\) |
| \([\ce{Cl^-}]\) | \(0.200\ \mathrm{mol\,L^{-1}}\) |

Notice that the density of the solution never entered the calculation. We already knew the final volume from the volumetric flask.

## When the concentration is given as a mass percentage

A mass percentage requires a different conversion.

Consider a solution labelled \(4.00\%\ \mathrm{m/m}\) calcium chloride.

This means that every \(100.0\ \mathrm{g}\) of solution contains \(4.00\ \mathrm{g}\) of calcium chloride.

It does not mean that every \(100.0\ \mathrm{mL}\) contains \(4.00\ \mathrm{g}\).

To find the amount concentration, you need three pieces of information:

- The mass percentage of the salt.
- The molar mass of the salt.
- The density of the solution.

The density connects the mass of the solution to its volume:

\[
\rho=\frac{m_{\text{solution}}}{V_{\text{solution}}}
\]

Rearranging:

\[
V_{\text{solution}}=
\frac{m_{\text{solution}}}{\rho}
\]

Once you have the volume, you can calculate the salt concentration in \(\mathrm{mol\,L^{-1}}\).

[IMAGE: A clear comparison of a 100.0 g sample of aqueous calcium chloride solution and its corresponding volume at a stated density of 1.04 g/mL. Highlight that 4.00 g is the mass of dissolved CaCl2 and 96.00 g is the remaining solution mass. Show the calculated solution volume as approximately 96.2 mL, making the distinction between mass percentage and volume-based concentration visually clear.]

### Worked example: Calculate ion concentrations from a mass percentage

An aqueous calcium chloride solution has a concentration of \(4.00\%\ \mathrm{m/m}\) and a density of \(1.04\ \mathrm{g\,mL^{-1}}\).

Calculate the concentrations of calcium ions and chloride ions.

Assume the salt is anhydrous and completely dissociated. The supplied density applies at the temperature of the solution.

**Step 1**: Choose a convenient mass of solution.

Use \(100.0\ \mathrm{g}\) of solution.

Because the concentration is \(4.00\%\ \mathrm{m/m}\), this sample contains:

\[
m(\ce{CaCl2})=4.00\ \mathrm{g}
\]

**Step 2**: Calculate the amount of calcium chloride.

The molar mass is:

\[
\begin{aligned}
M(\ce{CaCl2})
&=40.08+2(35.45)\\
&=110.98\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
n(\ce{CaCl2})
&=\frac{m}{M}\\
&=\frac{4.00}{110.98}\\
&=0.0360\ \mathrm{mol}
\end{aligned}
\]

**Step 3**: Calculate the volume of the solution.

This is where the density becomes necessary.

\[
\begin{aligned}
V&=\frac{m}{\rho}\\
&=\frac{100.0}{1.04}\\
&=96.15\ \mathrm{mL}
\end{aligned}
\]

Convert to litres:

\[
V=0.09615\ \mathrm{L}
\]

**Step 4**: Calculate the amount concentration of calcium chloride.

\[
\begin{aligned}
c_{\ce{CaCl2}}
&=\frac{n}{V}\\
&=\frac{0.0360}{0.09615}\\
&=0.375\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

**Step 5**: Calculate the individual ion concentrations.

The dissociation equation is:

\[
\ce{CaCl2(aq) -> Ca^{2+}(aq) + 2Cl^{-}(aq)}
\]

Therefore:

\[
[\ce{Ca^{2+}}]=0.375\ \mathrm{mol\,L^{-1}}
\]

\[
[\ce{Cl^-}]=2(0.375)
=0.750\ \mathrm{mol\,L^{-1}}
\]

The final answers are:

- Calcium ions: \(0.375\ \mathrm{mol\,L^{-1}}\).
- Chloride ions: \(0.750\ \mathrm{mol\,L^{-1}}\).

The key point is that the \(4.00\%\) mass percentage was not itself the molar concentration. We had to convert the mass of salt into moles and the mass of solution into volume.

If a question supplies a mass percentage but gives no density or other information that establishes the solution volume, you generally cannot determine its molar concentration uniquely.

## Three mistakes to avoid

There are a few tempting shortcuts that produce incorrect answers.

### 1. Confusing ionic charge with the number of ions

Consider magnesium chloride:

\[
\ce{MgCl2 -> Mg^{2+} + 2Cl^-}
\]

A common mistake is to multiply the magnesium chloride concentration by two for both ions.

But look carefully at the equation.

- There is one \(\ce{Mg^{2+}}\) ion per formula unit.
- There are two \(\ce{Cl^-}\) ions per formula unit.

The \(2+\) on magnesium describes its electrical charge. It does not mean that two magnesium ions are produced.

Therefore:

\[
[\ce{Mg^{2+}}]=c_{\ce{MgCl2}}
\]

\[
[\ce{Cl^-}]=2c_{\ce{MgCl2}}
\]

The coefficient tells you how many ions are produced. The superscript tells you the charge on each ion.

### 2. Forgetting the brackets around polyatomic ions

Consider calcium nitrate:

\[
\ce{Ca(NO3)2(aq) -> Ca^{2+}(aq) + 2NO3^-(aq)}
\]

The subscript 2 outside the brackets multiplies the entire nitrate group.

One formula unit produces two nitrate ions, not two nitrogen atoms or two oxygen atoms.

Consequently:

\[
[\ce{Ca^{2+}}]=c_{\ce{Ca(NO3)2}}
\]

\[
[\ce{NO3^-}]=2c_{\ce{Ca(NO3)2}}
\]

Treat polyatomic ions as complete groups when counting ions.

### 3. Using the wrong molar mass for a hydrated salt

Some ionic salts contain water molecules within their crystal structure. These are called hydrates.

For example:

\[
\ce{CuSO4.5H2O}
\]

This is copper(II) sulfate pentahydrate.

When it dissolves, each formula unit produces one copper ion and one sulfate ion. The five water molecules do not produce additional copper or sulfate ions.

If you are calculating the amount of salt from its mass, however, you must include the five water molecules when calculating the molar mass.

For example:

\[
\begin{aligned}
M(\ce{CuSO4.5H2O})
={}&63.55+32.07\\
&+4(16.00)+5(18.02)\\
={}&249.72\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

Using the molar mass of anhydrous \(\ce{CuSO4}\) would overestimate the number of moles present in a weighed sample of the hydrate.

The water changes the mass per formula unit, but not the ratio of copper ions to sulfate ions.

## When the simple multiplication rule stops working

So far, we have assumed that the salt dissolves completely and that the ions remain in solution.

That assumption is important.

Suppose you dissolve a soluble salt in water. The dissociation equation predicts the concentrations of the ions produced.

But what happens if you then add another solution that reacts with one of those ions?

The original ion concentration calculation is no longer enough to determine the final concentrations.

For example:

\[
\ce{BaCl2(aq) + Na2SO4(aq) ->
BaSO4(s) + 2NaCl(aq)}
\]

Barium sulfate forms a solid precipitate.

The reaction removes barium ions and sulfate ions from the dissolved solution. Sodium and chloride ions remain dissolved.

This means you must first calculate the amounts of the ions supplied by each salt, then account for the reaction before calculating the final concentrations.

The shortcut of multiplying salt concentration by an ion coefficient gives the concentrations immediately after dissolution, provided complete dissociation applies. It does not automatically give the concentrations after further chemical reactions.

This is an important distinction when you move from straightforward ion concentration questions to precipitation calculations.

[IMAGE: A physical representation of two clear solutions being combined in a conical flask: one contains dissolved barium chloride and the other dissolved sodium sulfate. Show a white barium sulfate precipitate forming and settling at the bottom, with sodium and chloride ions remaining dispersed in the liquid. Label the reactants, the solid product, and the spectator ions. Do not show calculated final concentrations.]

## Questions and solutions

Try each question before reading its solution. The questions move from direct stoichiometric calculations to problems involving concentration conversions, experimental preparation, and reactions between dissolved salts.

### Question 1

A solution of potassium carbonate, \(\ce{K2CO3}\), has an amount concentration of \(0.150\ \mathrm{mol\,L^{-1}}\).

Assuming complete dissociation, calculate:

a) The potassium-ion concentration.

b) The carbonate-ion concentration.

### Solution 1

The potassium-ion concentration is \(0.300\ \mathrm{mol\,L^{-1}}\), and the carbonate-ion concentration is \(0.150\ \mathrm{mol\,L^{-1}}\).

Start by writing the dissociation equation:

\[
\ce{K2CO3(aq) -> 2K+(aq) + CO3^{2-}(aq)}
\]

There are two potassium ions and one carbonate ion per formula unit.

For potassium ions:

\[
\begin{aligned}
[\ce{K+}]&=2(0.150)\\
&=0.300\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

For carbonate ions:

\[
\begin{aligned}
[\ce{CO3^{2-}}]&=1(0.150)\\
&=0.150\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

The important detail is that the subscript 2 in \(\ce{K2CO3}\) tells you how many potassium ions are produced. The \(2-\) charge on carbonate does not change the number of carbonate ions.

### Question 2

A student takes \(25.0\ \mathrm{mL}\) of a \(0.240\ \mathrm{mol\,L^{-1}}\) aluminium chloride solution and dilutes it to a final volume of \(200.0\ \mathrm{mL}\) using a volumetric flask.

Calculate the final concentrations of:

a) Aluminium ions.

b) Chloride ions.

Assume complete dissociation and no other reactions.

[IMAGE: An annotated dilution apparatus showing a pipette transferring exactly 25.0 mL of aluminium chloride stock solution into a 200.0 mL volumetric flask. Show distilled water being added to the calibration mark, with the original stock concentration and final flask volume labelled. Do not show the calculated final ion concentrations.]

### Solution 2

The final aluminium-ion concentration is \(0.0300\ \mathrm{mol\,L^{-1}}\), and the final chloride-ion concentration is \(0.0900\ \mathrm{mol\,L^{-1}}\).

There are two separate calculations: first account for dilution, then use the dissociation equation.

**Step 1**: Calculate the amount of aluminium chloride in the original sample.

Use:

\[
n=cV
\]

Convert the initial volume into litres:

\[
25.0\ \mathrm{mL}=0.0250\ \mathrm{L}
\]

Therefore:

\[
\begin{aligned}
n(\ce{AlCl3})
&=0.240\times0.0250\\
&=0.00600\ \mathrm{mol}
\end{aligned}
\]

**Step 2**: Calculate the new salt concentration.

The dilution changes the volume, but the amount of aluminium chloride remains unchanged.

\[
\begin{aligned}
c_{\text{final}}
&=\frac{n}{V_{\text{final}}}\\
&=\frac{0.00600}{0.2000}\\
&=0.0300\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

**Step 3**: Write the dissociation equation.

\[
\ce{AlCl3(aq) -> Al^{3+}(aq) + 3Cl^-(aq)}
\]

**Step 4**: Calculate both ion concentrations.

For aluminium ions:

\[
[\ce{Al^{3+}}]=0.0300\ \mathrm{mol\,L^{-1}}
\]

For chloride ions:

\[
\begin{aligned}
[\ce{Cl^-}]&=3(0.0300)\\
&=0.0900\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

The final answers are:

- \([\ce{Al^{3+}}]=0.0300\ \mathrm{mol\,L^{-1}}\).
- \([\ce{Cl^-}]=0.0900\ \mathrm{mol\,L^{-1}}\).

A useful check is to remember that dilution does not change the ratio of ions produced by the salt. The chloride concentration must remain three times the aluminium-ion concentration.

### Question 3

An aqueous solution contains \(14.8\ \mathrm{g\,L^{-1}}\) of anhydrous magnesium nitrate, \(\ce{Mg(NO3)2}\).

Calculate:

a) The amount concentration of magnesium nitrate.

b) The magnesium-ion concentration.

c) The nitrate-ion concentration.

Use a molar mass of \(148.3\ \mathrm{g\,mol^{-1}}\) for magnesium nitrate.

Assume complete dissociation.

### Solution 3

The magnesium-ion concentration is \(0.0998\ \mathrm{mol\,L^{-1}}\), and the nitrate-ion concentration is \(0.200\ \mathrm{mol\,L^{-1}}\).

The given concentration is a mass concentration, not an amount concentration. You must convert it before using the dissociation equation.

**Step 1**: Convert the mass concentration into amount concentration.

Use:

\[
c_{\text{salt}}=
\frac{\text{mass concentration}}{M}
\]

Substitute:

\[
\begin{aligned}
c_{\ce{Mg(NO3)2}}
&=\frac{14.8}{148.3}\\
&=0.0998\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

**Step 2**: Write the dissociation equation.

\[
\ce{Mg(NO3)2(aq) ->
Mg^{2+}(aq) + 2NO3^-(aq)}
\]

One formula unit produces one magnesium ion and two nitrate ions.

**Step 3**: Calculate the magnesium-ion concentration.

\[
\begin{aligned}
[\ce{Mg^{2+}}]
&=1(0.0998)\\
&=0.0998\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

**Step 4**: Calculate the nitrate-ion concentration.

\[
\begin{aligned}
[\ce{NO3^-}]
&=2(0.0998)\\
&=0.1996\\
&\approx0.200\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

The final answers are:

- \([\ce{Mg^{2+}}]=0.0998\ \mathrm{mol\,L^{-1}}\).
- \([\ce{NO3^-}]=0.200\ \mathrm{mol\,L^{-1}}\).

The trap was using \(14.8\) directly as though it were a molar concentration. Always check the units before applying the ion ratio.

### Question 4

An aqueous aluminium chloride solution is labelled \(4.00\%\ \mathrm{m/m}\). Its density is \(1.10\ \mathrm{g\,mL^{-1}}\).

Calculate the concentrations of aluminium ions and chloride ions.

Explain why interpreting the concentration as \(4.00\ \mathrm{g}\) of aluminium chloride per \(100\ \mathrm{mL}\) of solution would be incorrect.

Use a molar mass of \(133.33\ \mathrm{g\,mol^{-1}}\) for anhydrous aluminium chloride.

[IMAGE: A comparison illustrating mass percentage and density for an aluminium chloride solution. Show a 100.0 g solution sample containing 4.00 g AlCl3 and a separate graduated volume representation corresponding to 90.9 mL at the supplied density. Clearly distinguish the total solution mass from its volume.]

### Solution 4

The aluminium-ion concentration is \(0.330\ \mathrm{mol\,L^{-1}}\), and the chloride-ion concentration is \(0.990\ \mathrm{mol\,L^{-1}}\).

The density allows us to convert the mass of the solution into its volume. Without it, the molar concentration could not be determined from the mass percentage alone.

**Step 1**: Choose a \(100.0\ \mathrm{g}\) sample of solution.

A \(4.00\%\ \mathrm{m/m}\) solution contains:

\[
m(\ce{AlCl3})=4.00\ \mathrm{g}
\]

**Step 2**: Calculate the amount of aluminium chloride.

\[
\begin{aligned}
n(\ce{AlCl3})
&=\frac{m}{M}\\
&=\frac{4.00}{133.33}\\
&=0.0300\ \mathrm{mol}
\end{aligned}
\]

**Step 3**: Calculate the volume of the solution.

Use the density:

\[
V=\frac{m}{\rho}
\]

\[
\begin{aligned}
V&=\frac{100.0}{1.10}\\
&=90.91\ \mathrm{mL}\\
&=0.09091\ \mathrm{L}
\end{aligned}
\]

**Step 4**: Calculate the amount concentration of aluminium chloride.

\[
\begin{aligned}
c_{\ce{AlCl3}}
&=\frac{0.0300}{0.09091}\\
&=0.330\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

**Step 5**: Calculate the ion concentrations.

The dissociation equation is:

\[
\ce{AlCl3(aq) -> Al^{3+}(aq) + 3Cl^-(aq)}
\]

Therefore:

\[
[\ce{Al^{3+}}]=0.330\ \mathrm{mol\,L^{-1}}
\]

\[
[\ce{Cl^-}]=3(0.330)
=0.990\ \mathrm{mol\,L^{-1}}
\]

The original interpretation of \(4.00\ \mathrm{g}\) per \(100\ \mathrm{mL}\) is incorrect because the percentage is based on the mass of the entire solution, not its volume.

The actual volume of our \(100.0\ \mathrm{g}\) sample is approximately \(90.91\ \mathrm{mL}\), not \(100.0\ \mathrm{mL}\).

### Question 5

A student prepares \(100.0\ \mathrm{mL}\) of solution by dissolving \(3.22\ \mathrm{g}\) of sodium sulfate decahydrate, \(\ce{Na2SO4.10H2O}\), in water.

The student then mixes this solution with \(100.0\ \mathrm{mL}\) of \(0.0800\ \mathrm{mol\,L^{-1}}\) barium chloride solution.

A white precipitate forms.

The student calculates the concentrations immediately after mixing, before accounting for any reaction:

\[
[\ce{Na+}]=0.100\ \mathrm{mol\,L^{-1}}
\]

\[
[\ce{SO4^{2-}}]=0.0500\ \mathrm{mol\,L^{-1}}
\]

\[
[\ce{Ba^{2+}}]=0.0400\ \mathrm{mol\,L^{-1}}
\]

\[
[\ce{Cl^-}]=0.0800\ \mathrm{mol\,L^{-1}}
\]

They point out that the total positive and negative charges are balanced, so they claim these must be the final concentrations.

**Is the student correct?**

Calculate the actual final concentrations of all four ions after the reaction.

Use the following assumptions:

- The molar mass of \(\ce{Na2SO4.10H2O}\) is \(322.20\ \mathrm{g\,mol^{-1}}\).
- The solution volumes are additive, giving a final volume of \(200.0\ \mathrm{mL}\).
- Barium sulfate precipitation is quantitative, and its small equilibrium solubility is ignored.
- Sodium and chloride ions remain dissolved, and no other reactions occur.

[IMAGE: A two-stage chemical illustration. On the left, a 100.0 mL solution prepared from 3.22 g Na2SO4.10H2O and a separate 100.0 mL solution of 0.0800 mol/L BaCl2. On the right, show the solutions combined in one vessel, with a white BaSO4 precipitate settled at the bottom and dissolved ions remaining in the liquid. Label only the supplied quantities, reactants, and precipitate. Do not show final ion concentrations.]

### Solution 5

No. The student's concentrations are correct immediately after mixing if no reaction occurs, but they are not the final concentrations.

The actual final concentrations are:

| Ion | Final concentration |
|---|---:|
| \(\ce{Na+}\) | \(0.100\ \mathrm{mol\,L^{-1}}\) |
| \(\ce{Cl^-}\) | \(0.0800\ \mathrm{mol\,L^{-1}}\) |
| \(\ce{SO4^{2-}}\) | \(0.0100\ \mathrm{mol\,L^{-1}}\) |
| \(\ce{Ba^{2+}}\) | Approximately \(0\ \mathrm{mol\,L^{-1}}\) |

The mistake is subtle: the student's charge balance is correct, but they have ignored the chemical reaction that removes ions from solution.

**Step 1**: Calculate the amount of sodium sulfate supplied by the hydrate.

The hydrate contains one sodium sulfate formula unit per formula unit of the crystal.

Its molar mass includes the ten water molecules:

\[
M(\ce{Na2SO4.10H2O})
=322.20\ \mathrm{g\,mol^{-1}}
\]

Therefore:

\[
\begin{aligned}
n(\ce{Na2SO4.10H2O})
&=\frac{3.22}{322.20}\\
&=0.0100\ \mathrm{mol}
\end{aligned}
\]

This provides:

\[
n(\ce{Na+})=2(0.0100)
=0.0200\ \mathrm{mol}
\]

and:

\[
n(\ce{SO4^{2-}})=0.0100\ \mathrm{mol}
\]

The ten water molecules contribute to the mass of the hydrate, but they do not produce additional sodium or sulfate ions.

**Step 2**: Calculate the amount of barium chloride added.

Use:

\[
n=cV
\]

The volume is \(0.1000\ \mathrm{L}\).

\[
\begin{aligned}
n(\ce{BaCl2})
&=0.0800\times0.1000\\
&=0.00800\ \mathrm{mol}
\end{aligned}
\]

The dissociation equation is:

\[
\ce{BaCl2(aq) -> Ba^{2+}(aq) + 2Cl^-(aq)}
\]

Therefore:

\[
n(\ce{Ba^{2+}})=0.00800\ \mathrm{mol}
\]

\[
n(\ce{Cl^-})=2(0.00800)
=0.0160\ \mathrm{mol}
\]

**Step 3**: Identify the reaction between the ions.

Barium ions react with sulfate ions to form solid barium sulfate:

\[
\ce{Ba^{2+}(aq) + SO4^{2-}(aq) ->
BaSO4(s)}
\]

The mole ratio is \(1:1\).

We initially have:

| Ion | Initial amount |
|---|---:|
| \(\ce{Ba^{2+}}\) | \(0.00800\ \mathrm{mol}\) |
| \(\ce{SO4^{2-}}\) | \(0.0100\ \mathrm{mol}\) |

Barium ions are the limiting reactant because there are fewer moles of them.

All \(0.00800\ \mathrm{mol}\) of barium ions react with \(0.00800\ \mathrm{mol}\) of sulfate ions.

**Step 4**: Calculate the amounts remaining after precipitation.

For barium ions:

\[
n_{\text{remaining}}(\ce{Ba^{2+}})=0
\]

For sulfate ions:

\[
\begin{aligned}
n_{\text{remaining}}(\ce{SO4^{2-}})
&=0.0100-0.00800\\
&=0.00200\ \mathrm{mol}
\end{aligned}
\]

The sodium and chloride ions are spectator ions. They do not participate in the precipitation reaction, so their amounts remain unchanged.

We therefore have:

| Ion | Final amount |
|---|---:|
| \(\ce{Na+}\) | \(0.0200\ \mathrm{mol}\) |
| \(\ce{Cl^-}\) | \(0.0160\ \mathrm{mol}\) |
| \(\ce{SO4^{2-}}\) | \(0.00200\ \mathrm{mol}\) |
| \(\ce{Ba^{2+}}\) | Approximately \(0\ \mathrm{mol}\) |

**Step 5**: Calculate the final concentrations.

The total solution volume is:

\[
V=100.0+100.0=200.0\ \mathrm{mL}
\]

\[
V=0.2000\ \mathrm{L}
\]

For sodium ions:

\[
\begin{aligned}
[\ce{Na+}]
&=\frac{0.0200}{0.2000}\\
&=0.100\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

For chloride ions:

\[
\begin{aligned}
[\ce{Cl^-}]
&=\frac{0.0160}{0.2000}\\
&=0.0800\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

For sulfate ions:

\[
\begin{aligned}
[\ce{SO4^{2-}}]
&=\frac{0.00200}{0.2000}\\
&=0.0100\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

For barium ions, the assumed quantitative precipitation gives an approximately zero concentration.

**Step 6**: Check electrical neutrality.

The total positive charge concentration is:

\[
\begin{aligned}
[\text{positive charge}]
&=[\ce{Na+}]+2[\ce{Ba^{2+}}]\\
&=0.100+0\\
&=0.100
\end{aligned}
\]

The total negative charge concentration is:

\[
\begin{aligned}
[\text{negative charge}]
&=[\ce{Cl^-}]+2[\ce{SO4^{2-}}]\\
&=0.0800+2(0.0100)\\
&=0.100
\end{aligned}
\]

The charges balance.

But here is the important conceptual point: **charge balance alone does not prove that you have accounted for every reaction.**

The student's original concentrations also satisfied charge balance. Both barium and sulfate are divalent ions, and the precipitation reaction removes equal numbers of positive and negative charges from solution.

Consequently, even a perfectly balanced charge calculation can miss a reaction that removes ions.

You must account for the chemistry first, then use charge balance as a check.

## Choosing the next calculation

The method in this article answers one particular question: given the concentration of a salt, what concentrations of ions does it produce?

Keep the following decision guide in mind:

| If the question gives you... | Your first move |
|---|---|
| \(\mathrm{mol\,L^{-1}}\) of salt | Multiply by the ion's coefficient |
| \(\mathrm{g\,L^{-1}}\) of salt | Divide by molar mass, then multiply by the coefficient |
| \(\%\ \mathrm{m/m}\) | Use density to convert solution mass into volume |
| A hydrated salt's mass | Use the hydrate's full molar mass |
| A diluted salt solution | Calculate the new salt concentration before finding the ion concentrations |
| A mixture in which ions react | Calculate the initial amounts, account for the reaction, then divide remaining amounts by final volume |

Once you can calculate ion concentrations reliably, you can use them in other HSC Chemistry calculations.

For example, if you are working with hydrogen or hydroxide ion concentrations, the next step may be calculating pH or pOH. See [pH, pOH and Ion Concentrations for HSC Chemistry](https://tutorgum.com/learn/resources/hsc-chemistry-ph-poh-and-ion-concentrations/).

If the ion concentration is unknown and you need to determine it experimentally using reaction volumes and mole ratios, the problem changes. That is where [Precipitation Titrations: Calculate an Unknown Concentration](https://tutorgum.com/learn/resources/hsc-chemistry-precipitation-titrations/) becomes useful.

The distinction is simple: calculating ion concentrations from a known salt concentration uses the dissociation equation; determining an unknown concentration from experimental reaction data requires you to work backwards through the reaction's mole ratio.