How to Calculate Final Ion Concentrations When Mixing Salts

You mix 100.0 mL of 0.200 mol L\(^{-1}\) sodium chloride with 150.0 mL of 0.100 mol L\(^{-1}\) calcium chloride. Both salts dissolve, and no reaction occurs.

What is the final concentration of chloride ions, \(\mathrm{Cl^-}\)?

A tempting approach is to average the two starting concentrations. That gives 0.150 mol L\(^{-1}\). However, it is incorrect. The solutions have different volumes, and each formula unit of calcium chloride produces two chloride ions, not one.

The correct answer is 0.200 mol L\(^{-1}\).

To calculate final ion concentrations after mixing nonreacting salt solutions, follow one central rule:

Calculate the moles of each ion supplied by every salt, add the contributions from all sources, then divide by the total final volume.

This works because, when no reaction occurs, the amount of each dissolved ion is conserved. Its concentration can change through dilution, but its total number of moles does not.

01What actually happens when you mix two salt solutions?

Imagine two separate solutions on a laboratory bench.

The first contains sodium chloride, \(\mathrm{NaCl}\). The second contains calcium chloride, \(\mathrm{CaCl_2}\).

In water, both salts dissociate into ions:

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

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

Before mixing, the ions are already separated and surrounded by water molecules. When the solutions are combined, these ions become distributed throughout the same volume of solution.

Nothing consumes the chloride ions. Instead, chloride ions from both original solutions contribute to the final mixture.

At the particle level, the important change is that ions from both solutions now occupy the same mixture. The chloride ions have two different sources, but they are chemically identical.

Why can we conserve the moles of each ion?

Think of the moles as a count of how much of each type of ion is present.

If you pour two containers of water into one larger container, the total amount of water is unchanged. The volume occupied by the water changes, but you haven’t magically created or destroyed any water.

The same conservation idea applies to the dissolved ions in a nonreacting mixture.

For example:

• Sodium ions from \(\mathrm{NaCl}\) remain sodium ions.
• Calcium ions from \(\mathrm{CaCl_2}\) remain calcium ions.
• Chloride ions from both salts remain chloride ions.

The analogy has one important limitation: ions can react with other substances. If a precipitate forms, an acid-base reaction occurs, or a complex ion forms, the amount of the original dissolved ion may change. You must account for that reaction before calculating the remaining concentrations.

For this article, assume all salts are soluble, fully dissociated, and nonreacting.

02The method: calculate each ion separately

There are four main steps.

Step 1: Calculate the moles of each salt

Start with the concentration equation:

\[
n=cV
\]

Where:

• \(n\) is the amount of salt, in mol.
• \(c\) is the concentration of the salt solution, in mol L\(^{-1}\).
• \(V\) is the volume of the solution, in L.

Convert every volume from millilitres to litres before substituting.

For example:

\[
100.0\text{ mL}=0.1000\text{ L}
\]

Remember that the concentration of a salt solution describes the amount of its formula units per litre. It does not necessarily equal the concentration of each individual ion.

Step 2: Work out how many moles of each ion the salt produces

Use the dissociation equation to identify the number of each ion produced per formula unit.

For example:

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

One mole of calcium chloride produces:

• 1 mol of \(\mathrm{Ca^{2+}}\)
• 2 mol of \(\mathrm{Cl^-}\)

Therefore, if you have 0.0150 mol of \(\mathrm{CaCl_2}\), you have:

\[
n(\mathrm{Ca}^{2+})=0.0150\text{ mol}
\]

\[
n(\mathrm{Cl}^{-})=2(0.0150)=0.0300\text{ mol}
\]

The subscript matters. Forgetting it is one of the easiest ways to lose marks in these calculations.

Step 3: Add the moles of identical ions

If two or more salts produce the same ion, add their contributions.

For example, if one solution supplies 0.0200 mol of chloride ions and another supplies 0.0300 mol, the total is:

\[
n_{\text{total}}(\mathrm{Cl}^{-})
=0.0200+0.0300
\]

\[
=0.0500\text{ mol}
\]

It doesn’t matter which salt supplied the chloride ions. Once dissolved, all chloride ions are the same chemical species.

Step 4: Divide by the final total volume

The final concentration is:

\[
[\text{ion}]_{\text{final}}
=
\frac{n_{\text{total}}(\text{ion})}
{V_{\text{final}}}
\]

For two solutions with additive volumes:

\[
V_{\text{final}}=V_1+V_2
\]

Use the total volume of the mixture, not the original volume of either solution.

For more precise experimental preparations, the final volume may be measured directly or established using a volumetric flask. If the question supplies that final volume, use it instead of assuming the original volumes are additive.

04The tricky part: when two salts produce the same positive ion

So far, we’ve combined salts that share a negative ion.

What happens if they share a positive ion instead?

This is where students often make a mistake. They calculate the final concentration of each salt separately, then forget that both salts contribute to the concentration of the shared ion.

Consider sodium chloride and sodium sulfate.

Their dissociation equations are:

\[
\ce{NaCl -> Na+ + Cl-}
\]

\[
\ce{Na2SO4 -> 2Na+ + SO4^2-}
\]

Both produce sodium ions, but sodium sulfate produces twice as many sodium ions per formula unit.

05Mixing salt solutions versus diluting one salt solution

These two processes look similar, but they require slightly different bookkeeping.

Single-solution dilution

Suppose you take 100.0 mL of 0.100 mol L\(^{-1}\) sodium chloride and add 100.0 mL of water.

The original amount of sodium chloride is:

\[
n=0.100\times0.1000
=0.0100\text{ mol}
\]

After adding water, the final volume is 0.2000 L.

Therefore:

\[
c_{\text{final}}=\frac{0.0100}{0.2000}
\]

\[
=0.0500\text{ mol L}^{-1}
\]

Both sodium and chloride concentrations are now 0.0500 mol L\(^{-1}\).

No extra salt has been introduced.

Mixing two different salt solutions

Now replace the water with 100.0 mL of 0.100 mol L\(^{-1}\) sodium sulfate.

The final volume is still 0.2000 L, but the second salt supplies additional sodium ions.

The final concentrations are:

• \([\mathrm{Na}^{+}]=0.150\text{ mol L}^{-1}\)
• \([\mathrm{Cl}^{-}]=0.0500\text{ mol L}^{-1}\)
• \([\mathrm{SO_{4}}^{2-}]=0.0500\text{ mol L}^{-1}\)

The important distinction is that dilution changes the concentration of existing ions, while mixing can introduce additional sources of the same ion.

You must account for both effects.

06A useful check: conservation of electrical charge

There is another way to check whether your final ion concentrations are reasonable.

An ordinary solution containing dissolved salts must be electrically neutral overall.

That means the total positive charge concentration must equal the total negative charge concentration.

For the sodium chloride and calcium chloride example:

\[
[\mathrm{Na}^{+}]+2[\mathrm{Ca}^{2+}]
=
[\mathrm{Cl}^{-}]
\]

Substituting:

\[
0.0800+2(0.0600)=0.200
\]

\[
0.200=0.200
\]

The charges balance.

Notice that calcium ions have a charge of \(+2\), so their contribution must be multiplied by two. Similarly, sulfate ions have a charge of \(-2\).

This is a useful arithmetic check, particularly when a mixture contains several salts.

However, charge balance is not a replacement for calculating the moles of each ion. It checks your answer; it does not tell you which salt supplied each ion.

07Common mistakes to avoid

A reliable decision process is:

1. Identify every salt in the mixture.
2. Calculate the moles of each salt using \(n=cV\).
3. Write its dissociation equation.
4. Calculate the moles of each ion.
5. Add contributions from all sources of the same ion.
6. Determine the final solution volume.
7. Divide each total ion amount by that volume.
8. Check electrical charge balance where appropriate.

All volumes must be expressed in litres when using concentration in mol L\(^{-1}\).

The worked examples and questions below assume complete dissociation, no reaction, and additive volumes unless a different final volume is explicitly supplied. All numerical values are illustrative teaching data.

08Questions and solutions

Try each question before reading its solution. The questions move from straightforward mixing towards situations where you must combine dilution, stoichiometry, and reverse calculations.

Question 1

50.0 mL of 0.200 mol L\(^{-1}\) potassium chloride, \(\mathrm{KCl}\), is mixed with 150.0 mL of 0.0500 mol L\(^{-1}\) sodium nitrate, \(\mathrm{NaNO_3}\).

Assuming no reaction occurs and the volumes are additive, calculate the final concentrations of:

a) \(\mathrm{K^+}\)

b) \(\mathrm{Cl^-}\)

c) \(\mathrm{Na^+}\)

d) \(\mathrm{NO_3^-}\)

Solution 1

The final concentrations are 0.0500 mol L\(^{-1}\) for both potassium and chloride ions, and 0.0375 mol L\(^{-1}\) for both sodium and nitrate ions.

Step 1: Calculate the moles of each salt.

For potassium chloride:

\[
n=0.200\times0.0500
=0.0100\text{ mol}
\]

For sodium nitrate:

\[
n=0.0500\times0.1500
=0.00750\text{ mol}
\]

Step 2: Identify the ions produced.

\[
\ce{KCl -> K+ + Cl-}
\]

Therefore:

\[
n(\mathrm{K}^{+})=n(\mathrm{Cl}^{-})=0.0100\text{ mol}
\]

For sodium nitrate:

\[
\ce{NaNO3 -> Na+ + NO3-}
\]

Therefore:

\[
n(\mathrm{Na}^{+})=n(\ce{NO3-})=0.00750\text{ mol}
\]

Step 3: Calculate the final volume.

\[
V_{\text{final}}=0.0500+0.1500
\]

\[
=0.2000\text{ L}
\]

Step 4: Calculate the final concentrations.

For potassium and chloride:

\[
[\mathrm{K}^{+}]=[\mathrm{Cl}^{-}]
\]

\[
=\frac{0.0100}{0.2000}
\]

\[
=0.0500\text{ mol L}^{-1}
\]

For sodium and nitrate:

\[
[\mathrm{Na}^{+}]=[\ce{NO3-}]
\]

\[
=\frac{0.00750}{0.2000}
\]

\[
=0.0375\text{ mol L}^{-1}
\]

The final answers are:

Each ion has only one source, so no additional ion contributions need to be combined.

Question 2

75.0 mL of 0.200 mol L\(^{-1}\) sodium chloride is mixed with 125.0 mL of 0.0800 mol L\(^{-1}\) sodium sulfate.

Assume no reaction occurs.

a) Calculate the final concentration of sodium ions.

b) Calculate the final concentrations of chloride and sulfate ions.

c) Explain why the final sodium concentration is not equal to the final concentration of either individual salt.

Solution 2

The final concentrations are 0.175 mol L\(^{-1}\) for sodium ions, 0.0750 mol L\(^{-1}\) for chloride ions, and 0.0500 mol L\(^{-1}\) for sulfate ions.

The key is that both salts supply sodium ions.

Step 1: Calculate the moles of each salt.

For sodium chloride:

\[
n=0.200\times0.0750
=0.0150\text{ mol}
\]

For sodium sulfate:

\[
n=0.0800\times0.1250
=0.0100\text{ mol}
\]

Step 2: Calculate the moles of each ion.

From sodium chloride:

\[
n(\mathrm{Na}^{+})=0.0150\text{ mol}
\]

\[
n(\mathrm{Cl}^{-})=0.0150\text{ mol}
\]

From sodium sulfate:

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

\[
n(\mathrm{SO_{4}}^{2-})=0.0100\text{ mol}
\]

Step 3: Combine the sodium contributions.

\[
n_{\text{total}}(\mathrm{Na}^{+})
=0.0150+0.0200
\]

\[
=0.0350\text{ mol}
\]

Step 4: Calculate the final volume.

\[
V_{\text{final}}=0.0750+0.1250
\]

\[
=0.2000\text{ L}
\]

Step 5: Calculate the final concentrations.

Sodium:

\[
[\mathrm{Na}^{+}]=\frac{0.0350}{0.2000}
\]

\[
=0.175\text{ mol L}^{-1}
\]

Chloride:

\[
[\mathrm{Cl}^{-}]=\frac{0.0150}{0.2000}
\]

\[
=0.0750\text{ mol L}^{-1}
\]

Sulfate:

\[
[\mathrm{SO_{4}}^{2-}]=\frac{0.0100}{0.2000}
\]

\[
=0.0500\text{ mol L}^{-1}
\]

Step 6: Check charge balance.

The total positive charge concentration is:

\[
0.175
\]

The total negative charge concentration is:

\[
0.0750+2(0.0500)=0.175
\]

The charges balance.

The tempting mistake is to calculate sodium from just one of the salts. However, sodium sulfate contributes two sodium ions per formula unit, so both contributions must be included.

Question 3

100.0 mL of 0.150 mol L\(^{-1}\) magnesium nitrate, \(\mathrm{Mg(NO_3)_2}\), is mixed with 200.0 mL of 0.0500 mol L\(^{-1}\) potassium chloride.

Assume no reaction occurs.

a) Calculate the final concentration of magnesium ions.

b) Calculate the final concentration of nitrate ions.

c) Calculate the final concentrations of potassium and chloride ions.

d) Explain why the nitrate concentration is twice the concentration of magnesium nitrate after mixing.

Solution 3

The final concentrations are:

• \([\mathrm{Mg}^{2+}]=0.0500\text{ mol L}^{-1}\)
• \([\ce{NO3-}]=0.100\text{ mol L}^{-1}\)
• \([\mathrm{K}^{+}]=0.0333\text{ mol L}^{-1}\)
• \([\mathrm{Cl}^{-}]=0.0333\text{ mol L}^{-1}\)

The important detail is the two nitrate ions produced by every formula unit of magnesium nitrate.

Step 1: Calculate the moles of magnesium nitrate.

\[
n=0.150\times0.1000
\]

\[
=0.0150\text{ mol}
\]

Step 2: Calculate the moles of magnesium and nitrate ions.

The dissociation equation is:

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

Therefore:

\[
n(\mathrm{Mg}^{2+})=0.0150\text{ mol}
\]

\[
n(\ce{NO3-})=2(0.0150)
\]

\[
=0.0300\text{ mol}
\]

Step 3: Calculate the moles of potassium chloride.

\[
n=0.0500\times0.2000
\]

\[
=0.0100\text{ mol}
\]

Since:

\[
\ce{KCl -> K+ + Cl-}
\]

we have:

\[
n(\mathrm{K}^{+})=n(\mathrm{Cl}^{-})=0.0100\text{ mol}
\]

Step 4: Calculate the final volume.

\[
V_{\text{final}}=0.1000+0.2000
\]

\[
=0.3000\text{ L}
\]

Step 5: Calculate all final concentrations.

Magnesium:

\[
[\mathrm{Mg}^{2+}]=\frac{0.0150}{0.3000}
\]

\[
=0.0500\text{ mol L}^{-1}
\]

Nitrate:

\[
[\ce{NO3-}]=\frac{0.0300}{0.3000}
\]

\[
=0.100\text{ mol L}^{-1}
\]

Potassium:

\[
[\mathrm{K}^{+}]=\frac{0.0100}{0.3000}
\]

\[
=0.0333\text{ mol L}^{-1}
\]

Chloride:

\[
[\mathrm{Cl}^{-}]=\frac{0.0100}{0.3000}
\]

\[
=0.0333\text{ mol L}^{-1}
\]

The nitrate concentration is twice the magnesium ion concentration because each magnesium nitrate formula unit produces two nitrate ions.

This ratio remains true after dilution because both ions occupy the same final volume.

Question 4

A student mixes 80.0 mL of 0.200 mol L\(^{-1}\) potassium chloride with 120.0 mL of 0.100 mol L\(^{-1}\) magnesium chloride.

The mixture is transferred quantitatively into a 250.0 mL volumetric flask. Deionised water is added until the bottom of the meniscus reaches the calibration mark.

a) Calculate the final concentrations of potassium, magnesium, and chloride ions.

b) A student uses 200.0 mL as the final volume because this is the sum of the original solution volumes. Explain why this is incorrect.

Solution 4

The final concentrations are 0.0640 mol L\(^{-1}\) potassium ions, 0.0480 mol L\(^{-1}\) magnesium ions, and 0.160 mol L\(^{-1}\) chloride ions.

The crucial detail is that the final volume is 250.0 mL, not 200.0 mL. The volumetric flask establishes the actual final volume.

Step 1: Calculate the moles of potassium chloride.

\[
n=0.200\times0.0800
\]

\[
=0.0160\text{ mol}
\]

Therefore:

\[
n(\mathrm{K}^{+})=0.0160\text{ mol}
\]

\[
n(\mathrm{Cl}^{-})=0.0160\text{ mol}
\]

Step 2: Calculate the moles of magnesium chloride.

\[
n=0.100\times0.1200
\]

\[
=0.0120\text{ mol}
\]

Using:

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

we obtain:

\[
n(\mathrm{Mg}^{2+})=0.0120\text{ mol}
\]

\[
n(\mathrm{Cl}^{-})=2(0.0120)
\]

\[
=0.0240\text{ mol}
\]

Step 3: Add the chloride contributions.

\[
n_{\text{total}}(\mathrm{Cl}^{-})
=0.0160+0.0240
\]

\[
=0.0400\text{ mol}
\]

Step 4: Use the actual final volume.

\[
V_{\text{final}}=0.2500\text{ L}
\]

The extra water added to reach the calibration mark increases the final volume beyond the combined starting volumes.

Step 5: Calculate the final concentrations.

Potassium:

\[
[\mathrm{K}^{+}]=\frac{0.0160}{0.2500}
\]

\[
=0.0640\text{ mol L}^{-1}
\]

Magnesium:

\[
[\mathrm{Mg}^{2+}]=\frac{0.0120}{0.2500}
\]

\[
=0.0480\text{ mol L}^{-1}
\]

Chloride:

\[
[\mathrm{Cl}^{-}]=\frac{0.0400}{0.2500}
\]

\[
=0.160\text{ mol L}^{-1}
\]

Step 6: Check charge balance.

\[
[\mathrm{K}^{+}]+2[\mathrm{Mg}^{2+}]
\]

\[
=0.0640+2(0.0480)
\]

\[
=0.160
\]

This equals the chloride concentration.

The student’s mistake is assuming that the initial volumes always determine the final volume. That assumption is only appropriate when the question specifies or permits additive volumes. Here, the volumetric flask explicitly establishes a different final volume.

Question 5

A chemist prepares a solution by mixing three salt solutions:

The solutions are mixed without reacting and the final mixture is made up to exactly 300.0 mL.

Once the mixture is homogeneous, a 30.0 mL aliquot is removed using a pipette and transferred into a second volumetric flask. Deionised water is added until the final volume reaches 120.0 mL.

The chloride concentration in this diluted sample is found to be 0.0325 mol L\(^{-1}\).

a) Calculate the chloride concentration in the original 300.0 mL mixture.

b) Determine the initial concentration of the magnesium chloride solution.

c) Calculate the final concentrations of all five ions in the original mixture: \(\mathrm{Na^+}\), \(\mathrm{SO_4^{2-}}\), \(\mathrm{K^+}\), \(\mathrm{Mg^{2+}}\), and \(\mathrm{Cl^-}\).

d) Explain why the fact that only one-tenth of the original mixture was sampled does not mean you should multiply the measured chloride concentration by ten.

Solution 5

The original mixture has a chloride concentration of 0.130 mol L\(^{-1}\), and the initial magnesium chloride concentration was 0.0967 mol L\(^{-1}\).

The final concentrations in the original mixture are:

This question combines two different ideas: dilution of an aliquot and conservation of individual ion amounts.

Step 1: Work backwards from the diluted sample.

The original mixture has a volume of 300.0 mL.

A 30.0 mL aliquot is diluted to 120.0 mL.

The dilution factor is:

\[
\frac{120.0}{30.0}=4.00
\]

Therefore, the original chloride concentration was four times the measured concentration:

\[
[\mathrm{Cl}^{-}]_{\text{original}}
=0.0325\times4.00
\]

\[
=0.130\text{ mol L}^{-1}
\]

Step 2: Calculate the total moles of chloride in the original mixture.

Use the original final volume of 0.3000 L.

\[
n(\mathrm{Cl}^{-})=cV
\]

\[
=0.130\times0.3000
\]

\[
=0.0390\text{ mol}
\]

The original mixture therefore contained 0.0390 mol of chloride ions.

Step 3: Calculate the chloride supplied by the known solutions.

From sodium sulfate:

\[
\ce{Na2SO4 -> 2Na+ + SO4^2-}
\]

There is no chloride in this salt, so:

\[
n(\mathrm{Cl}^{-})=0
\]

From potassium chloride:

\[
n(\ce{KCl})=0.200\times0.0500
\]

\[
=0.0100\text{ mol}
\]

Since each formula unit produces one chloride ion:

\[
n(\mathrm{Cl}^{-})=0.0100\text{ mol}
\]

Step 4: Determine the chloride supplied by magnesium chloride.

The total chloride amount is 0.0390 mol.

The known potassium chloride contribution is 0.0100 mol.

Therefore:

\[
n(\mathrm{Cl}^{-})_{\text{from MgCl2}}
=0.0390-0.0100
\]

\[
=0.0290\text{ mol}
\]

Step 5: Calculate the moles of magnesium chloride.

The dissociation equation is:

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

Each mole of magnesium chloride produces two moles of chloride ions.

Therefore:

\[
n(\ce{MgCl2})=\frac{0.0290}{2}
\]

\[
=0.0145\text{ mol}
\]

Step 6: Determine the original magnesium chloride concentration.

The original magnesium chloride solution had a volume of 150.0 mL, or 0.1500 L.

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

\[
=\frac{0.0145}{0.1500}
\]

\[
=0.0967\text{ mol L}^{-1}
\]

Step 7: Calculate the concentrations of the remaining ions.

The original mixture has a final volume of 0.3000 L.

First, consider sodium sulfate.

\[
n(\ce{Na2SO4})=0.100\times0.1000
\]

\[
=0.0100\text{ mol}
\]

Each formula unit produces two sodium ions:

\[
n(\mathrm{Na}^{+})=0.0200\text{ mol}
\]

Therefore:

\[
[\mathrm{Na}^{+}]=\frac{0.0200}{0.3000}
\]

\[
=0.0667\text{ mol L}^{-1}
\]

The sulfate amount is:

\[
n(\mathrm{SO_{4}}^{2-})=0.0100\text{ mol}
\]

Therefore:

\[
[\mathrm{SO_{4}}^{2-}]=\frac{0.0100}{0.3000}
\]

\[
=0.0333\text{ mol L}^{-1}
\]

Next, potassium chloride supplies:

\[
n(\mathrm{K}^{+})=0.0100\text{ mol}
\]

Therefore:

\[
[\mathrm{K}^{+}]=\frac{0.0100}{0.3000}
\]

\[
=0.0333\text{ mol L}^{-1}
\]

Finally, magnesium chloride supplies:

\[
n(\mathrm{Mg}^{2+})=0.0145\text{ mol}
\]

Therefore:

\[
[\mathrm{Mg}^{2+}]=\frac{0.0145}{0.3000}
\]

\[
=0.0483\text{ mol L}^{-1}
\]

The chloride concentration was already determined:

\[
[\mathrm{Cl}^{-}]=0.130\text{ mol L}^{-1}
\]

Step 8: Check electrical charge balance.

The total positive charge concentration is:

\[
[\mathrm{Na}^{+}]+[\mathrm{K}^{+}]+2[\mathrm{Mg}^{2+}]
\]

Substituting:

\[
=0.0667+0.0333+2(0.0483)
\]

\[
=0.1966\text{ mol L}^{-1}
\]

The total negative charge concentration is:

\[
[\mathrm{Cl}^{-}]+2[\mathrm{SO_{4}}^{2-}]
\]

Substituting:

\[
=0.130+2(0.0333)
\]

\[
=0.1966\text{ mol L}^{-1}
\]

The charges balance, confirming that the calculated ion concentrations are consistent.

Step 9: Understand the aliquot trap.

The tempting mistake is to multiply the measured chloride concentration by ten because 30.0 mL is one-tenth of the original 300.0 mL mixture.

But that factor of ten relates to the amount of solution sampled, not its concentration.

Before dilution, the 30.0 mL aliquot has exactly the same ion concentrations as the original homogeneous mixture. It simply contains one-tenth of the total moles of each ion.

When that aliquot is diluted to 120.0 mL, its volume increases by a factor of four, so its concentration decreases by a factor of four.

That is why you multiply the measured concentration by four, not ten, to recover the original concentration.

This distinction between the amount sampled and the concentration of the sample is the key to solving the problem correctly.

09Where to go next

You now have a reliable method for calculating ion concentrations when nonreacting salt solutions are mixed: conserve the moles of each ion, combine contributions from shared ions, and divide by the actual final volume.

The next important distinction is recognising when this method no longer applies directly.

If you mix an acid and a base, for example, hydrogen ions and hydroxide ions react to form water. You must calculate the reaction first, determine which species remain, and only then calculate the final concentrations and pH.

For that situation, see pH After Dilution and Acid-Base Mixing | HSC Chemistry.

For the relationship between hydrogen ion concentration, hydroxide ion concentration, pH, and pOH, see pH, pOH and Ion Concentrations for HSC Chemistry.

The decision to make before any calculation is simple: are the original ions still present unchanged, or has a chemical reaction altered the mixture? Getting that right tells you whether you can use direct ion conservation or need to do reaction stoichiometry first.