Aqueous ions and concentration: keep amount and volume separate

MCAT content library / Chemistry & physics / Aqueous ions and concentration: keep amount and volume separate

Track each species in moles before dividing by the final volume. Use dissociation coefficients and charge balance to check the result.

About 14 minutes, including the worked examples and checks.

From solutions to buffers · Lesson 2 of 6 · Continue in this path

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• Concept: a dissolved amount is not yet a concentration
• Pattern: read the denominator as carefully as the solute
• Application: prepare two solutions, then mix measured portions
• Rule: dilution conserves solute amount; mixing may require more bookkeeping
• Audit the result with charge balance
• Check your reasoning
• Continue learning
• Sources

Concept: a dissolved amount is not yet a concentration

A solution calculation has two separate questions: what chemical species are present, and how much solution contains them? AAMC category 5A includes dissolved ions, hydration, hydronium, and concentration units.¹ Keeping species, amount, and final volume in separate columns prevents several different errors from producing the same plausible-looking number.

An aqueous ion is surrounded by water molecules. Water’s polarity allows ion–dipole attractions that help stabilize separated ions. When a soluble ionic compound dissociates, its formula determines the relative amounts of cations and anions. A strong electrolyte produces ions essentially completely under the stated solution conditions; a dissolved molecular substance may remain uncharged or react only partly to form ions.²

For a fully dissociated calcium chloride solution, write CaCl₂ → Ca²⁺ + 2 Cl⁻. One mole of dissolved formula units supplies one mole of calcium ions and two moles of chloride ions. The subscript two belongs to chloride’s count; the superscript 2+ belongs to calcium’s charge. They perform different jobs.

Likewise, the polyatomic ion NH₄⁺ has one positive charge on the whole ammonium ion. It is not four separate hydrogen ions. H₃O⁺, hydronium, represents a proton associated with water. In introductory aqueous equations, H⁺ is often shorthand for this hydrated proton rather than an isolated bare proton in solution.³

Pattern: read the denominator as carefully as the solute

Molarity is amount of solute divided by final solution volume: c = n/V, with n in mol and V in L. Thus 0.100 mol in 0.500 L is 0.200 mol/L, written 0.200 M. The denominator is not the volume of water initially poured into the flask. Dissolving a solid or mixing liquids can change the final volume.⁴

Unit or label	Meaning	Information needed for conversion
mol/L, or M	Moles of stated solute per liter of solution	Species identity and final solution volume
g/L	Grams of stated solute per liter of solution	Molar mass to convert to mol/L
% by mass	Grams of solute per 100 g of solution	Solution density to connect mass and volume
% mass/volume	Grams of solute per 100 mL of solution	The explicit mass/volume convention

For example, 5.85 g/L of NaCl is 0.100 mol/L when its supplied molar mass is 58.5 g/mol. A 5.85% mass/volume solution contains 5.85 g per 100 mL, which is ten times as much per liter. “Percent” without a stated basis is incomplete information; mass percent and mass/volume percent need not be numerically equal.⁵

Application: prepare two solutions, then mix measured portions

The following preparation and mixing data are illustrative. Assume pure anhydrous salts, complete dissolution and dissociation, no precipitation or other reaction, and the stated final volumes. Use CaCl₂ = 111 g/mol and NaCl = 58.5 g/mol.

Stock	Preparation	Dissolved amount	Formula-unit concentration
A: CaCl2	2.22 g brought to 100.0 mL final volume	0.0200 mol	0.200 M
B: NaCl	1.17 g brought to 200.0 mL final volume	0.0200 mol	0.100 M

The stocks contain the same amount of salt formula units, but stock A has half the volume. Its formula-unit concentration is therefore twice as large. In A, [Ca²⁺] = 0.200 M and [Cl⁻] = 0.400 M. In B, both [Na⁺] and [Cl⁻] equal 0.100 M. Brackets here indicate the molar concentration of the particular species inside them.

Now combine 50.0 mL of A with 150.0 mL of B. For this example, the measured final volume is 200.0 mL. Calculate each contribution as n = cV before adding it to the common vessel.

Ion	From 50.0 mL A	From 150.0 mL B	Total divided by 0.2000 L
Ca2+	0.0100 mol	0	0.0500 M
Na+	0	0.0150 mol	0.0750 M
Cl−	0.0200 mol	0.0150 mol	0.175 M

Chloride has two sources. Adding the original chloride concentrations, 0.400 M + 0.100 M, would incorrectly treat both concentrations as though they referred to the final volume. Even their unweighted average would be wrong because the sampled volumes differ. The general nonreacting mixture relation is c_final = (c_AV_A + c_BV_B)/V_final for the same species.

Rule: dilution conserves solute amount; mixing may require more bookkeeping

Suppose 40.0 mL of the mixture is transferred and diluted to 200.0 mL. The transferred portion contains 0.175 mol/L × 0.0400 L = 0.00700 mol of chloride. Dilution leaves that amount unchanged while increasing its volume fivefold, so [Cl⁻] becomes 0.0350 M. Calcium and sodium concentrations also fall by a factor of five.

The familiar relation c₁V₁ = c₂V₂ is this conservation statement written twice. It applies to the amount actually transferred, not automatically to the entire stock bottle. It also assumes no gain, loss, or reaction of the tracked solute during dilution.⁴

Adding 200 mL of water is different from bringing a sample to 200 mL final volume. If the final volume is not supplied and volume additivity is not justified, an exact concentration cannot be obtained from component volumes alone. The bookkeeping still gives the amount; the missing quantity is the denominator.

Audit the result with charge balance

For the mixture above, positive charge equivalents per liter are 2[Ca²⁺] + [Na⁺] = 2(0.0500) + 0.0750 = 0.175. Negative charge equivalents are [Cl⁻] = 0.175. The charges balance even though there are more chloride ions than calcium and sodium ions combined. Electrical neutrality requires equal total positive and negative charge, not equal numbers of positive and negative ions.

This check can expose a missing coefficient or counterion. It cannot prove that every concentration is correct: multiplying all three concentrations by the same wrong factor would preserve charge balance. Pair it with independent amount and volume checks. If a new mixture precipitates a salt, forms a complex, or transfers protons, first account for that chemistry; the assumption of independently conserved free-ion amounts has then changed.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

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From solutions to buffers

Lesson 2 of 6.

• Previous in From solutions to buffers: Moles, formulas, and yield: conserve the reacting material
• Next in From solutions to buffers: Acid–base strength: distinguish concentration from dissociation
• Return to this path’s overview

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From solutions to buffers

Build a quantitative acid–base explanation and apply it to amino acid charge.

1. Moles, formulas, and yield: conserve the reacting material
2. Aqueous ions and concentration: keep amount and volume separate · You are here
3. Acid–base strength: distinguish concentration from dissociation
4. Weak acids, bases, and salts: choose a defensible pH calculation
5. Buffers: explain pH resistance and its limits
6. Amino acid and peptide charge: count the groups that can change

Related lessons

• Moles, formulas, and yield: conserve the reacting material
• Acid–base strength: distinguish concentration from dissociation
• Weak acids, bases, and salts: choose a defensible pH calculation

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Sources

1. AAMC — Content Category 5A

2. OpenStax Chemistry 2e — Electrolytes

3. OpenStax Chemistry 2e — Brønsted–Lowry Acids and Bases

4. OpenStax Chemistry 2e — Molarity

5. OpenStax Chemistry 2e — Other Units for Solution Concentrations