How to Calculate the Total Number of Ions in a Dissolved Salt

A student dissolves 5.85 g of sodium chloride in water and calculates that the sample contains \(6.02\times10^{22}\) particles. Their calculation is correct, but their answer is incomplete. They have counted the NaCl formula units, not the individual ions produced when the salt dissolves.

Each NaCl formula unit produces two ions: one \(\mathrm{Na^+}\) ion and one \(\mathrm{Cl^-}\) ion. The solution therefore contains \(1.20\times10^{23}\) ions in total.

The method is straightforward once you keep track of exactly what you’re counting:

1. Convert the mass of salt into moles.
2. Convert moles into the number of formula units using Avogadro’s constant.
3. Multiply by the number of ions produced by each formula unit.

The third step is where students most often lose marks. Let’s work through why.

01The method: from mass of salt to number of ions

Before calculating anything, identify the difference between two quantities:

• Formula units: the number of simplest whole-number ionic ratios represented by the salt’s formula.
• Ions: the individual positively and negatively charged particles produced when the salt dissolves.

For example, sodium chloride has the formula \(\mathrm{NaCl}\). Its formula tells us that sodium and chloride ions occur in a 1:1 ratio.

When sodium chloride dissolves completely:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

One formula unit produces two ions.

Calcium chloride behaves differently:

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

One formula unit produces three ions: one calcium ion and two chloride ions.

This means that knowing the number of formula units is not quite enough. You must also know how many ions each formula unit produces.

The three calculations

Step 1: Calculate the amount of salt.

Use the relationship between mass, molar mass, and amount:

\[ n_{\text{salt}}=\frac{m}{M} \]

Where:

• \(n_{\text{salt}}\) is the amount of salt, in mol.
• \(m\) is the mass of salt dissolved, in g.
• \(M\) is the molar mass of the salt, in \(\mathrm{g\,mol^{-1}}\).

Step 2: Calculate the number of formula units.

Multiply the amount of salt by Avogadro’s constant:

\[ N_{\text{formula units}}=n_{\text{salt}}N_A \]

Avogadro’s constant is:

\[ N_A=6.022\times10^{23}\ \mathrm{mol^{-1}} \]

Step 3: Calculate the total number of ions.

Multiply the number of formula units by the number of ions produced per formula unit:

\[ \boxed{ N_{\text{ions}}=n_{\text{salt}}\times N_A\times \nu } \]

Here, \(\nu\) (Greek letter nu) represents the total number of ions produced by one formula unit.

Combining all three steps gives the most useful equation:

\[ \boxed{ N_{\text{ions}}=\frac{m}{M}\times N_A\times\nu } \]

This is the main relationship to remember.

Why the units work

Consider what happens when you substitute the quantities:

\[ \frac{\mathrm{g}}{\mathrm{g\,mol^{-1}}} \times \frac{\mathrm{formula\ units}}{\mathrm{mol}} \times \frac{\mathrm{ions}}{\mathrm{formula\ unit}} \]

The grams cancel, the moles cancel, and the formula units cancel.

The remaining unit is simply the number of ions.

This is a useful check. If your final answer is supposed to be a number of ions, it should not have units of mol/L or g.

02How to identify the number of ions per formula unit

The value of \(\nu\) comes from the dissociation equation.

You need to separate the salt into its positive and negative ions, then add their coefficients.

Consider these examples:

There are two important details here.

First, count polyatomic ions as single ions.

A nitrate ion, \(\mathrm{NO_3^-}\), contains one nitrogen atom and three oxygen atoms. However, the entire group carries one charge and behaves as one ion.

Therefore, one formula unit of magnesium nitrate produces three ions, not nine:

• One \(\mathrm{Mg^{2+}}\).
• Two \(\mathrm{NO_3^-}\).

Second, do not count charge as the number of ions.

One calcium ion has a \(2+\) charge, but it is still just one ion.

For calcium chloride, the total is three ions per formula unit, not four. The total charge magnitude is four, but the number of individual ions is three.

05An important distinction: more ions per formula unit does not always mean more ions per gram

Here’s a comparison that exposes a common mistake.

Suppose you have 1.00 g of sodium chloride and 1.00 g of calcium chloride. Both samples dissolve completely.

Which sample produces more ions?

You might predict calcium chloride because each formula unit produces three ions, compared with two for sodium chloride.

But that prediction ignores molar mass.

Sodium chloride actually produces more ions from the same mass.

Why?

Calcium chloride produces more ions per formula unit, but each formula unit is also heavier. A 1.00 g sample therefore contains fewer calcium chloride formula units.

For a fixed mass of salt, the relationship is:

\[ N_{\text{ions}}\propto\frac{\nu}{M} \]

This gives you a useful decision rule:

• If two samples contain equal amounts in mol, compare their ions per formula unit.
• If two samples have equal masses, consider both their molar masses and their ions per formula unit.

Never compare salts using the number of ions per formula unit alone when their masses are different.

06What happens in an actual dissolution experiment?

So far, we have calculated ion numbers from a known mass. But how would you investigate this in a laboratory?

The difficulty is that you cannot directly count \(10^{22}\) individual ions. Instead, you measure a quantity that can be determined accurately and use chemical relationships to infer the number of ions.

A suitable experimental arrangement would use:

1. An analytical balance and a weighing boat to measure the mass of dry salt.
2. A beaker containing distilled water, into which the salt is transferred.
3. A glass stirring rod to help the salt dissolve.
4. A volumetric flask if you also need to prepare a solution with a known final volume.

What should the experiment show?

If you dissolve a soluble ionic salt in water, you expect the solution to contain mobile charged particles.

A conductivity probe can provide evidence for this. A solution containing mobile ions conducts electricity, unlike pure distilled water, which has very low conductivity.

However, there’s a limitation.

A conductivity reading does not directly tell you the exact number of ions present. Conductivity also depends on:

• The concentration of the ions.
• The charges carried by the ions.
• How quickly the ions move through the solution.
• The temperature of the solution.

Two solutions containing different numbers of ions can produce similar conductivity readings.

The better method is therefore to use the measured mass and the salt’s dissociation equation to calculate the theoretical ion count. Conductivity can support the conclusion that mobile ions are present, but it cannot replace the stoichiometric calculation.

This method also has assumptions. The calculation must account for the amount that actually dissolves, rather than automatically counting the entire mass placed in the beaker.

07Important assumptions and common mistakes

The equation

\[ N_{\text{ions}}=\frac{m}{M}\times N_A\times\nu \]

works directly when the salt dissolves completely and dissociates completely.

Here are the situations that can change your calculation.

A closer look at hydrated salts

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

For example, copper(II) sulfate pentahydrate has the formula:

\[ \mathrm{CuSO_4\cdot5H_2O} \]

When it dissolves:

\[ \mathrm{CuSO_4\cdot5H_2O(s)} \]

\[ \mathrm{\rightarrow Cu^{2+}(aq)+SO_4^{2-}(aq)+5H_2O(l)} \]

Each formula unit produces two ions.

The five water molecules become part of the solvent. They do not become ions.

This means you must include the five water molecules when calculating the molar mass, but you must not add five to the number of ions produced.

This distinction becomes particularly important in exam questions where the mass of a hydrated salt is provided.

08Questions and solutions

The following questions build from a standard mass-to-ion calculation towards more demanding problems involving mixtures, hydrates, and precipitation.

Question 1

A student dissolves 2.34 g of sodium chloride, \(\mathrm{NaCl}\), completely in water.

Given that the molar mass of sodium chloride is \(58.5\ \mathrm{g\,mol^{-1}}\), calculate the total number of ions produced.

Solution 1

The total number of ions is:

\[ \boxed{2.41\times10^{22}\text{ ions}} \]

First, calculate the amount of sodium chloride:

\[ n=\frac{m}{M} \]

\[ n=\frac{2.34}{58.5} \]

\[ n=0.0400\ \mathrm{mol} \]

Convert this into the number of formula units:

\[ N=0.0400\times6.022\times10^{23} \]

\[ N=2.41\times10^{22} \]

Sodium chloride dissociates according to:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

Each formula unit produces two ions.

Therefore:

\[ N_{\text{ions}}=2.41\times10^{22}\times2 \]

\[ \boxed{N_{\text{ions}}=4.82\times10^{22}} \]

The trap is stopping after calculating the number of formula units. The question asks for individual ions, so the result must be multiplied by two.

Question 2

A student dissolves 4.44 g of calcium chloride, \(\mathrm{CaCl_2}\), in water and prepares exactly 250.0 mL of solution.

The molar mass of calcium chloride is \(111.0\ \mathrm{g\,mol^{-1}}\).

Calculate:

a) The total number of ions in the solution.

b) The concentration of calcium ions.

c) The concentration of chloride ions.

Solution 2

The total number of ions is \(7.23\times10^{22}\), the calcium ion concentration is \(0.160\ \mathrm{mol\,L^{-1}}\), and the chloride ion concentration is \(0.320\ \mathrm{mol\,L^{-1}}\).

Step 1: Calculate the amount of calcium chloride.

\[ n=\frac{4.44}{111.0} \]

\[ n=0.0400\ \mathrm{mol} \]

Step 2: Determine the number of ions per formula unit.

The dissociation equation is:

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

Each formula unit produces three ions.

Step 3: Calculate the total number of ions.

\[ N_{\text{ions}}=0.0400\times6.022\times10^{23}\times3 \]

\[ \boxed{N_{\text{ions}}=7.23\times10^{22}} \]

Step 4: Calculate the concentration of calcium ions.

Each formula unit produces one calcium ion.

Therefore:

\[ n_{\mathrm{Ca^{2+}}}=0.0400\ \mathrm{mol} \]

Convert the final volume into litres:

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

Using \(c=n/V\):

\[ c_{\mathrm{Ca^{2+}}}=\frac{0.0400}{0.2500} \]

\[ \boxed{c_{\mathrm{Ca^{2+}}}=0.160\ \mathrm{mol\,L^{-1}}} \]

Step 5: Calculate the concentration of chloride ions.

Each formula unit produces two chloride ions.

Therefore:

\[ n_{\mathrm{Cl^-}}=2\times0.0400 \]

\[ n_{\mathrm{Cl^-}}=0.0800\ \mathrm{mol} \]

Now calculate the concentration:

\[ c_{\mathrm{Cl^-}}=\frac{0.0800}{0.2500} \]

\[ \boxed{c_{\mathrm{Cl^-}}=0.320\ \mathrm{mol\,L^{-1}}} \]

Notice that the chloride concentration is twice the calcium concentration.

The volume was needed to calculate concentration, but not the total number of ions. This is an important distinction when deciding which information to use in an exam question.

Question 3

A student completely dissolves 0.0800 mol of sodium sulfate, \(\mathrm{Na_2SO_4}\), and 0.0300 mol of potassium nitrate, \(\mathrm{KNO_3}\), in water.

The two solutions are then mixed. Assume that no precipitation or other reaction occurs.

Calculate:

a) The total amount, in mol, of ions present.

b) The total number of ions present.

c) The amount of each individual ion present, and identify which ion is most abundant.

Solution 3

The solution contains \(0.300\) mol of ions, equivalent to approximately \(1.81\times10^{23}\) ions.

The key is to calculate the contribution from each salt separately before adding them.

Step 1: Write the dissociation equation for sodium sulfate.

\[ \ce{Na2SO4(s) -> 2Na+(aq) + SO4^2-(aq)} \]

Each formula unit produces three ions.

For 0.0800 mol of sodium sulfate:

\[ n_{\text{ions}}=0.0800\times3 \]

\[ n_{\text{ions}}=0.240\ \mathrm{mol} \]

The individual ion amounts are:

\[ n_{\mathrm{Na^+}}=2\times0.0800=0.160\ \mathrm{mol} \]

\[ n_{\mathrm{SO_4^{2-}}}=0.0800\ \mathrm{mol} \]

Step 2: Write the dissociation equation for potassium nitrate.

\[ \ce{KNO3(s) -> K+(aq) + NO3-(aq)} \]

Each formula unit produces two ions.

For 0.0300 mol of potassium nitrate:

\[ n_{\text{ions}}=0.0300\times2 \]

\[ n_{\text{ions}}=0.0600\ \mathrm{mol} \]

The individual ion amounts are:

\[ n_{\mathrm{K^+}}=0.0300\ \mathrm{mol} \]

\[ n_{\mathrm{NO_3^-}}=0.0300\ \mathrm{mol} \]

Step 3: Add the total amounts of ions.

\[ n_{\text{total ions}}=0.240+0.0600 \]

\[ \boxed{n_{\text{total ions}}=0.300\ \mathrm{mol}} \]

Step 4: Convert the total amount into a number of ions.

\[ N=0.300\times6.022\times10^{23} \]

\[ \boxed{N=1.81\times10^{23}\text{ ions}} \]

Step 5: Compare the individual ion amounts.

Sodium ions are the most abundant, with 0.160 mol present.

The important reasoning is that the two salts do not necessarily produce equal numbers of ions per formula unit. Calculate each contribution using its own dissociation equation.

Question 4

A student dissolves 2.50 g of copper(II) sulfate pentahydrate, \(\mathrm{CuSO_4\cdot5H_2O}\), completely in water.

The final solution volume is 100.0 mL, and the molar mass of the hydrated salt is \(249.7\ \mathrm{g\,mol^{-1}}\).

Calculate:

a) The total number of aqueous ions produced.

b) The concentration of copper(II) ions.

c) The concentration of sulfate ions.

Explain why the five water molecules in the formula do not contribute to the total ion count.

Solution 4

The salt produces \(1.20\times10^{22}\) aqueous ions. Both the copper(II) ion and sulfate ion concentrations are \(0.100\ \mathrm{mol\,L^{-1}}\).

Step 1: Calculate the amount of hydrated copper(II) sulfate.

\[ n=\frac{2.50}{249.7} \]

\[ n=0.0100\ \mathrm{mol} \]

Step 2: Identify the ions produced.

The dissolution process is represented by:

\[ \ce{CuSO4.5H2O(s) -> Cu^2+(aq) + SO4^2-(aq) + 5H2O(l)} \]

Each formula unit produces two ions.

The five water molecules become part of the solvent. They are neutral molecules, not ions.

Therefore:

\[ \nu=2 \]

Step 3: Calculate the total number of ions.

\[ N=0.0100\times6.022\times10^{23}\times2 \]

\[ \boxed{N=1.20\times10^{22}\text{ ions}} \]

Step 4: Calculate the copper(II) ion concentration.

Each formula unit produces one copper(II) ion.

\[ n_{\mathrm{Cu^{2+}}}=0.0100\ \mathrm{mol} \]

Convert the volume:

\[ V=100.0\ \mathrm{mL}=0.1000\ \mathrm{L} \]

Therefore:

\[ c_{\mathrm{Cu^{2+}}}=\frac{0.0100}{0.1000} \]

\[ \boxed{c_{\mathrm{Cu^{2+}}}=0.100\ \mathrm{mol\,L^{-1}}} \]

Step 5: Calculate the sulfate ion concentration.

Each formula unit also produces one sulfate ion.

\[ n_{\mathrm{SO_4^{2-}}}=0.0100\ \mathrm{mol} \]

\[ c_{\mathrm{SO_4^{2-}}}=\frac{0.0100}{0.1000} \]

\[ \boxed{c_{\mathrm{SO_4^{2-}}}=0.100\ \mathrm{mol\,L^{-1}}} \]

The tempting mistake is to count the five water molecules as additional particles produced by dissociation. They are not ions, so they must not be included in the total ion count.

However, they are part of the hydrated salt’s formula and must be included when calculating its molar mass.

Question 5

Two separate solutions are prepared:

• Solution A contains 0.0200 mol of barium chloride, \(\mathrm{BaCl_2}\).
• Solution B contains an unknown amount of sodium sulfate, \(\mathrm{Na_2SO_4}\).

The solutions are mixed, producing a white precipitate of barium sulfate, \(\mathrm{BaSO_4}\).

The precipitate is filtered, washed, and dried. Its final mass is 3.50 g.

Assume:

• Both original salts dissolved and dissociated completely.
• The precipitation reaction proceeds to completion.
• Barium sulfate has negligible solubility under the conditions used.
• All sodium and chloride ions remain dissolved.
• No other reactions occur.

The molar mass of barium sulfate is \(233.4\ \mathrm{g\,mol^{-1}}\).

Calculate:

a) The original amount of sodium sulfate dissolved.

b) The total number of free ions remaining in the filtrate after precipitation.

c) The total number of free ions present immediately after the original salts dissociated, but before precipitation occurred.

Explain why your answers to parts (b) and (c) are different.

Solution 5

The original sodium sulfate amount is \(0.0150\) mol, and the filtrate contains \(4.52\times10^{22}\) free ions. Before precipitation, there were \(6.32\times10^{22}\) free ions.

The key is to distinguish between ions initially produced by dissolution and ions remaining freely dissolved after a chemical reaction.

Step 1: Use the precipitate mass to calculate the amount of barium sulfate formed.

First, calculate the amount of solid precipitate:

\[ n=\frac{m}{M} \]

\[ n=\frac{3.50}{233.4} \]

\[ n=0.0150\ \mathrm{mol} \]

Therefore, 0.0150 mol of barium sulfate was formed.

Step 2: Determine the original amount of sodium sulfate.

The net ionic equation is:

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

The mole ratio between barium ions and sulfate ions is 1:1.

Therefore, forming 0.0150 mol of barium sulfate requires 0.0150 mol of sulfate ions.

Each formula unit of sodium sulfate produces one sulfate ion:

\[ \ce{Na2SO4(s) -> 2Na+(aq) + SO4^2-(aq)} \]

Therefore:

\[ \boxed{n_{\mathrm{Na_2SO_4}}=0.0150\ \mathrm{mol}} \]

Step 3: Determine which ions remain after precipitation.

Before the reaction, barium chloride produced:

\[ n_{\mathrm{Ba^{2+}}}=0.0200\ \mathrm{mol} \]

\[ n_{\mathrm{Cl^-}}=2\times0.0200 \]

\[ n_{\mathrm{Cl^-}}=0.0400\ \mathrm{mol} \]

Sodium sulfate produced:

\[ n_{\mathrm{Na^+}}=2\times0.0150 \]

\[ n_{\mathrm{Na^+}}=0.0300\ \mathrm{mol} \]

\[ n_{\mathrm{SO_4^{2-}}}=0.0150\ \mathrm{mol} \]

The precipitation reaction consumes 0.0150 mol of barium ions and 0.0150 mol of sulfate ions.

Therefore, the remaining barium ion amount is:

\[ n_{\mathrm{Ba^{2+},remaining}}=0.0200-0.0150 \]

\[ n_{\mathrm{Ba^{2+},remaining}}=0.00500\ \mathrm{mol} \]

There are no sulfate ions remaining under the stated assumption of complete precipitation.

The sodium and chloride ions are spectators. They remain dissolved.

The final ion amounts are:

Step 4: Calculate the total amount of free ions remaining.

\[ n_{\text{total ions}}=0.00500+0.0300+0.0400 \]

\[ n_{\text{total ions}}=0.0750\ \mathrm{mol} \]

Convert into a number of ions:

\[ N=0.0750\times6.022\times10^{23} \]

\[ \boxed{N=4.52\times10^{22}\text{ ions}} \]

This is the number of free, dissolved ions remaining in the filtrate.

Step 5: Calculate the number of ions present before precipitation.

Immediately after dissolution, but before the solutions react, barium chloride produces three ions per formula unit:

\[ n_{\text{ions from BaCl}_2}=0.0200\times3 \]

\[ =0.0600\ \mathrm{mol} \]

Sodium sulfate also produces three ions per formula unit:

\[ n_{\text{ions from Na}_2\text{SO}_4}=0.0150\times3 \]

\[ =0.0450\ \mathrm{mol} \]

Add the contributions:

\[ n_{\text{initial ions}}=0.0600+0.0450 \]

\[ n_{\text{initial ions}}=0.1050\ \mathrm{mol} \]

Convert into a number of ions:

\[ N=0.1050\times6.022\times10^{23} \]

\[ \boxed{N=6.32\times10^{22}\text{ ions}} \]

Step 6: Explain the difference.

Initially, there were 0.1050 mol of free ions.

During precipitation, the reaction consumed:

• 0.0150 mol of \(\mathrm{Ba^{2+}}\) ions.
• 0.0150 mol of \(\mathrm{SO_4^{2-}}\) ions.

That is a total of 0.0300 mol of ions removed from the aqueous solution.

The remaining amount is:

\[ 0.1050-0.0300=0.0750\ \mathrm{mol} \]

This agrees with our calculation in part (b).

The tempting approach is to calculate the number of ions produced by each salt and simply add the results. That gives the initial count, but it ignores the subsequent precipitation reaction.

The barium and sulfate ions have not disappeared. They have become part of the solid barium sulfate ionic lattice. They are no longer free, dissolved ions in the filtrate.

This is the important distinction: the number of ions initially released by dissolution does not necessarily equal the number of free ions remaining after a chemical reaction.

A useful final check is charge balance. The remaining positive charge is:

\[ (0.00500\times2)+0.0300=0.0400 \]

The remaining negative charge is:

\[ 0.0400\times1=0.0400 \]

Both sides balance, supporting our result.

09Where to go next

Once you can calculate the number of ions produced by a dissolving salt, the next step is understanding what those ions do in solution.

In particular, precipitation reactions can remove ions from solution, while concentration calculations tell you how many moles of each ion remain per litre.

The distinction between total ions, individual ion concentrations, and ions removed through chemical reactions is central to solving more involved HSC Chemistry stoichiometry problems.