Solution Stoichiometry & Titrations

You’ll be able to

Relate molarity, moles, and volume, and calculate dilutions
Use titration data to find an unknown concentration
Apply the reaction mole ratio to titrations beyond the simple 1:1 case

Molarity connects moles to volume

Molarity (M) is moles of solute per liter of solution — the concentration currency of the lab. Rearranged, moles = molarity × volume (in liters), which lets you count particles just by measuring a volume of solution. Watch units: volumes given in mL must become liters (divide by 1000) before they enter the formula.

Molarity

M = mol ÷ L (so mol = M × V)

Volume must be in liters. 25.0 mL = 0.0250 L.

Titration: measuring the unknown

A titration slowly adds a solution of known concentration (the titrant) to a measured volume of unknown until they exactly react — the equivalence point, often flagged by an indicator’s color change. At that point the moles of titrant, combined with the balanced mole ratio, reveal the moles of unknown, and dividing by its volume gives its concentration. For a 1:1 acid–base reaction this collapses to the shortcut MₐVₐ = M_bV_b.

Titration relationships

mol titrant = M × V → mol unknown = mol titrant × (mole ratio) → M unknown = mol ÷ V

For a 1:1 reaction only, this simplifies to MₐVₐ = M_bV_b. When the ratio is not 1:1, you must apply it explicitly.

Worked example

A 25.0 mL sample of HCl is titrated with 0.100 M NaOH, reaching the equivalence point after 30.0 mL of NaOH. The reaction is HCl + NaOH → NaCl + H₂O. Find the concentration of the HCl.

1.Moles of NaOH added: M × V = 0.100 mol·L⁻¹ × 0.0300 L = 0.00300 mol.
2.Mole ratio is 1 HCl : 1 NaOH, so moles of HCl = 0.00300 mol.
3.Convert the HCl volume to liters: 25.0 mL = 0.0250 L.
4.Concentration = moles ÷ volume = 0.00300 mol ÷ 0.0250 L = 0.120 M.

Answer: 0.120 M HCl

Worked example

How many milliliters of 6.0 M HCl are needed to prepare 300 mL of 0.50 M HCl by dilution?

1.Dilution conserves moles of solute: M₁V₁ = M₂V₂.
2.Solve for V₁: V₁ = (M₂V₂) ÷ M₁ = (0.50 M × 300 mL) ÷ 6.0 M.
3.= 150 ÷ 6.0.
4.Because both volumes are in mL, no unit conversion is needed.

Answer: 25 mL of the 6.0 M stock (then add water to 300 mL)

Watch out

The MₐVₐ = M_bV_b shortcut only works when the acid and base react 1:1. For H₂SO₄ + 2NaOH or similar, you must include the mole ratio explicitly, or you will be off by exactly a factor of the ratio.

Checkpoint

It takes 20.0 mL of 0.50 M NaOH to neutralize 25.0 mL of HCl (1:1 reaction). What is the concentration of the HCl?

Checkpoint

A 25.0 mL sample of H₂SO₄ is titrated with 0.100 M NaOH, requiring 40.0 mL to reach the equivalence point. The reaction is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. What is the concentration of the H₂SO₄?

On the exam

Every solution-stoichiometry problem runs the same loop: convert volume + molarity to moles, cross the balanced equation with the mole ratio, then convert back using the target’s volume or molar mass. Keep volumes in liters throughout and the arithmetic stays clean.

Answer the 2 checkpoints as you read.