Solution Stoichiometry

12 min read Updated Mar 26, 2026

Solution stoichiometry is regular stoichiometry with one extra step: converting between volume and moles using molarity. If you can do grams-to-moles conversions and use mole ratios, you can do solution stoichiometry. The only new tool is the relationship M x V = moles.

The Key Relationship

Worked Example - Precipitation Reaction

Problem: How many mL of 0.200 M AgNO3 are needed to completely react with 35.0 mL of 0.100 M NaCl?

Balanced equation: AgNO3 + NaCl -> AgCl(s) + NaNO3

Step 1: Find moles of NaCl.

moles NaCl = M x V = (0.100 mol/L)(0.0350 L) = 0.00350 mol

Step 2: Use mole ratio.

1 mol AgNO3 : 1 mol NaCl
moles AgNO3 needed = 0.00350 mol

Step 3: Convert moles to volume.

V = moles / M = 0.00350 / 0.200 = 0.0175 L = 17.5 mL

Titration Calculations

A titration is a controlled reaction between a solution of known concentration (the titrant, in the buret) and a solution of unknown concentration (the analyte, in the flask). The goal is to find the unknown concentration.

Titration apparatus showing a buret filled with blue titrant solution positioned above an Erlenmeyer flask containing pink analyte solution, with a stopcock controlling the flow and a drop falling into the flask — A standard titration setup. The titrant (known concentration) is added drop by drop from the buret into the Erlenmeyer flask containing the analyte (unknown concentration). The color change at the equivalence point signals that stoichiometrically equal amounts have reacted. *Credit: Wikimedia Commons, CC BY 2.0*

At the equivalence point, the titrant has supplied exactly enough moles to react completely with the analyte. No excess of either remains.

Titration Worked Example

Problem: A 25.0 mL sample of HCl of unknown concentration is titrated with 0.150 M NaOH. The equivalence point is reached after 32.0 mL of NaOH is added. What is the concentration of HCl?

Balanced equation: HCl + NaOH -> NaCl + H2O (1:1 ratio)

Step 1: Find moles of NaOH used.

moles NaOH = (0.150)(0.0320) = 0.00480 mol

Step 2: Use mole ratio.

1 mol HCl : 1 mol NaOH
moles HCl = 0.00480 mol

Step 3: Find molarity of HCl.

M = moles / V = 0.00480 / 0.0250 = 0.192 M HCl

Non-1:1 Titrations

When the acid and base react in a ratio other than 1:1, you must account for stoichiometry.

Example: Titrating H2SO4 with NaOH.

Balanced equation: H2SO4 + 2 NaOH -> Na2SO4 + 2 H2O

Here, 1 mole of H2SO4 reacts with 2 moles of NaOH. At the equivalence point:

moles NaOH = 2 x moles H2SO4

If 20.0 mL of 0.100 M H2SO4 is titrated with 0.100 M NaOH:

moles H2SO4 = (0.100)(0.0200) = 0.00200 mol
moles NaOH needed = 2 x 0.00200 = 0.00400 mol
Volume NaOH = 0.00400 / 0.100 = 0.0400 L = 40.0 mL

Notice it took twice the volume of NaOH even though the molarities were equal - because each H2SO4 molecule requires two NaOH molecules.

Titration vs. Dilution - Do Not Confuse Them

Both use M x V, but for completely different reasons:

	Dilution	Titration
What happens	Add solvent to reduce concentration	Add a reactant to consume the analyte
Chemical reaction?	No	Yes
Equation	M1V1 = M2V2 (same solute)	M_aV_a = M_bV_b (different solutes reacting)
What stays constant	Moles of solute	Nothing - moles are consumed
Purpose	Prepare a less concentrated solution	Determine an unknown concentration

Back Titration

Sometimes the analyte does not react cleanly with an indicator or is insoluble. In a back titration, you add a known excess of one reagent, let it react with the analyte, and then titrate the leftover excess with a second standard solution.

Logic:

Moles of reagent added (known)
Moles of reagent that reacted with the analyte = moles added - moles left over
Use stoichiometry to find moles of analyte

This technique appears in MCAT passages about antacids (excess HCl + antacid, then titrate remaining HCl with NaOH) and calcium carbonate analysis.

What volume of 0.250 M KOH is needed to neutralize 40.0 mL of 0.125 M H2SO4?

Click to reveal answer

40.0 mL. H2SO4 + 2 KOH -> K2SO4 + 2 H2O. Moles H2SO4 = (0.125)(0.0400) = 0.00500 mol. Moles KOH needed = 2 x 0.00500 = 0.0100 mol. Volume KOH = 0.0100 / 0.250 = 0.0400 L = 40.0 mL. Even though H2SO4 has a lower molarity, the volumes are equal because each H2SO4 requires 2 KOH - the 2:1 ratio compensates for the concentration difference.

How does a titration calculation differ from a dilution calculation, even though both use M x V?

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Dilution preserves moles (no reaction); titration consumes moles (reaction occurs). In dilution, M1V1 = M2V2 because the same solute is simply spread through more volume. In titration, M_aV_a = M_bV_b (for 1:1 reactions) because the moles of acid are consumed by an equal number of moles of base. The equations look similar but describe fundamentally different processes.