Worked example 1
What volume of 0.250 M NaOH is needed to neutralise 25.0 mL of 0.100 M HCl?
Try it first: Write the balanced equation and check the mole ratio before calculating.
0 of 4 steps revealed.
Chemical ReactionsSolution StoichiometryContent level: Core 24 min
Solution Stoichiometry and Titration Calculations
What you'll be able to do: Use molarity and mole ratios to calculate concentrations, volumes and masses in solution reactions.
Introduction
Reactions in solution are measured with a burette rather than a balance, so molarity replaces molar mass as the way into moles. The stoichiometry itself does not change at all.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Convert between solution volume, molarity and moles
- Apply mole ratios to reactions carried out in solution
- Calculate an unknown concentration from titration data
- Distinguish the equivalence point from the endpoint
Lesson
Molarity as a conversion factor
Molarity is moles of solute per litre of solution, so it works in both directions. Multiply a volume in litres by molarity to get moles, or divide moles by molarity to get the volume needed. Treat it exactly as you treat molar mass in a mass-based problem.
n = M x V
The three-step route
Convert the given quantity to moles, apply the mole ratio from the balanced equation, then convert to whatever is requested. The only thing that changes between problems is which conversion sits at each end: molarity for solutions, molar mass for solids, and the ideal gas law for gases.
The mole ratio only ever acts on moles. Never apply coefficients directly to a volume or a molarity.
Titration in practice
A titrant of known concentration is delivered from a burette into a measured volume of analyte until the reaction is exactly complete. Recording the initial and final burette readings gives the volume delivered, which is the single measurement the whole calculation depends on.
M₁ V₁ / a = M₂ V₂ / b for a stoichiometric ratio a : b
Equivalence point versus endpoint
The equivalence point is the theoretical moment when the moles added exactly satisfy the stoichiometry. The endpoint is when the indicator changes colour. A well-chosen indicator makes them nearly coincide, and the small gap between them is the main systematic error in the technique.
A permanganate titration is self-indicating, because the first faint persistent pink shows the analyte is exhausted.
Ratios that are not one to one
The shortcut M₁V₁ = M₂V₂ is only valid when one mole of titrant reacts with one mole of analyte. With H₂SO₄ and NaOH the ratio is 1:2, and ignoring that halves or doubles the answer. Always write the balanced equation before reaching for a formula.
M₁V₁ = M₂V₂ is a dilution formula and a 1:1 titration shortcut. It is not a general titration equation.
Count acid and base equivalents before using M V
At equivalence, stoichiometric moles of acid and base have reacted; equal volumes or equal molarities are not required. A diprotic base such as Ca(OH)₂ supplies two OH⁻ equivalents per formula unit, so it needs twice as many HCl moles for complete neutralization. Write the balanced equation before turning molarity and volume into moles.
2 HCl + Ca(OH)₂ → CaCl₂ + 2 H₂O.
Key ideas
Definition
Molarity
Moles of solute per litre of solution, symbol M.
Definition
Titration
Adding a solution of known concentration until a reaction with an unknown is exactly complete.
Definition
Equivalence point
The point at which the added moles exactly satisfy the reaction stoichiometry.
Definition
Endpoint
The observed indicator change, used as an experimental estimate of the equivalence point.
Rule
Volume in litres
Molarity is defined per litre, so millilitre readings must be divided by 1000 before use.
Equation
Moles from molarity
n = M x V
- n = moles of solute
- M = molarity in mol per litre
- V = solution volume in litres
Equation
Titration relationship
n(titrant) × (b / a) = n(analyte)
- a = coefficient of the titrant in the balanced equation
- b = coefficient of the analyte in the balanced equation
Worked examples
Worked example 2
A 20.0 mL sample of H₂SO₄ requires 32.0 mL of 0.150 M NaOH to reach the equivalence point. What is the acid concentration?
Try it first: This is the case where the 1:1 shortcut fails.
0 of 4 steps revealed.
Worked example 3
What mass of AgCl precipitates when 50.0 mL of 0.200 M AgNO₃ is mixed with excess NaCl solution?
0 of 4 steps revealed.
Worked example 4
What volume of 0.100 M HCl neutralizes 25.0 mL of 0.0800 M Ca(OH)₂?
Try it first: Convert base volume to liters and use the 2:1 acid:base ratio.
0 of 3 steps revealed.
Common mistakes
Using millilitres directly in n = M x V.
Why it's wrong: Molarity is moles per litre, so a millilitre value inflates the answer by a factor of 1000.
Check instead: Divide every volume by 1000 before substituting.
Applying M₁V₁ = M₂V₂ to a titration with a 1:2 ratio.
Why it's wrong: That shortcut silently assumes a one-to-one stoichiometry.
Check instead: Write the balanced equation and route the calculation through moles.
Assuming the equivalence point is always at pH 7.
Why it's wrong: The salt formed can be acidic or basic when a weak acid or base is titrated.
Check instead: Only a strong acid with a strong base gives equivalence at pH 7.
Using the total mixed volume when calculating moles of a reactant.
Why it's wrong: Each reactant supplies moles based on its own volume before mixing.
Check instead: Use the individual volume for moles, and the combined volume only for a final concentration.
After the lesson · Question bank
Independent practice
Work through new problems on this topic on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
No questions are available for this topic yet. Explore the unit question bank.
Your next step
Lesson: Not marked completed
Marking a lesson completed records that you finished reading it. Practice establishes mastery separately. Saved on this device.
Continue through the unit’s lessons and available practice.
What you should now know
Molarity converts between volume and moles, so a solution stoichiometry problem is the familiar three-step route with molarity at each end. In a titration, a measured volume of known concentration is added until the equivalence point, and the mole ratio from the balanced equation then gives the unknown concentration. Careful unit conversion between millilitres and litres is where most errors occur.
- Molarity converts between solution volume and moles
- Route every calculation through moles before using the mole ratio
- Convert millilitres to litres every time
- M₁V₁ = M₂V₂ only applies to 1:1 stoichiometry
- The endpoint is the observed estimate of the true equivalence point
Sources and further reading
This lesson is original Chem Help content. No external sources were adapted.