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Titration to molarity and mass of solute

SCH3U Grade 11 Chemistry · Solutions and Solubility

A technician finds an unlabelled 250.0 mL volumetric flask of dilute hydrochloric acid in the stockroom. A 25.00 mL aliquot is pipetted out and titrated against standardised 0.1000 mol/L sodium hydroxide; the phenolphthalein endpoint arrives after 22.50 mL of titrant. Find the moles of base delivered, the concentration of the acid, the moles of HCl in the whole flask, and the mass of HCl it holds, taking the molar mass as 36.46 g/mol.

Step 1 of 4 · solve for Amount of solute

A burette reading is a volume, not an amount. Multiply it by the standardised concentration first — from here on the chain runs in moles.

Rearranged for n
n=C Vn = C \, V
Your values, in your units
n=(0.1 M) (22.5 mL)n = \left( 0.1\ \text{M} \right) \, \left( 22.5\ \text{mL} \right)
Converted to base units
n=(0.1 M) (0.0225 L)n = \left( 0.1\ \text{M} \right) \, \left( 0.0225\ \text{L} \right)
Answer
n=2.25n = 2.25

Carried onward at full precision, not this rounded figure.

Open the Molarity (C = n/V) solver →

Step 2 of 4 · solve for Molar concentration

HCl and NaOH react one to one, so those moles of base count as moles of acid. Divide by the pipetted 25.00 mL aliquot — never by the titre.

Rearranged for C
C=nVC = \frac{n}{V}
2.25 mmolcarried from step 1
Your values, in your units
C=(0.00225 mol)(25 mL)C = \frac{\left( 0.00225\ \text{mol} \right)}{\left( 25\ \text{mL} \right)}
Converted to base units
C=(0.00225 mol)(0.025 L)C = \frac{\left( 0.00225\ \text{mol} \right)}{\left( 0.025\ \text{L} \right)}
Answer
C=90C = 90

Carried onward at full precision, not this rounded figure.

Open the Molarity (C = n/V) solver →

Step 3 of 4 · solve for Amount of solute

Scale back up to the whole flask. The concentration is the same everywhere in it; only the volume changes.

Rearranged for n
n=C Vn = C \, V
90 mol/m³carried from step 2
Your values, in your units
n=(90 mol/m3) (250 mL)n = \left( 90\ \text{mol/m}^{3} \right) \, \left( 250\ \text{mL} \right)
Converted to base units
n=(0.09 M) (0.25 L)n = \left( 0.09\ \text{M} \right) \, \left( 0.25\ \text{L} \right)
Answer
n=22.5n = 22.5

Carried onward at full precision, not this rounded figure.

Open the Molarity (C = n/V) solver →

Step 4 of 4 · solve for Mass

Moles to grams, which is the form a stockroom label actually wants.

Rearranged for m
m=n Mm = n \, M
22.5 mmolcarried from step 3
Your values, in your units
m=(0.0225 mol) (36.46 g/mol)m = \left( 0.0225\ \text{mol} \right) \, \left( 36.46\ \text{g/mol} \right)
Answer
m=820m = 820

Carried onward at full precision, not this rounded figure.

Open the Moles from Mass (n = m/M) solver →

Why this order

A titration never measures a concentration directly. It measures a volume, and moles are the bridge — which is why step 1 converts the titre before anything else happens. Concentration times volume is the only honest thing a burette reading can become. The balanced equation then licenses the leap in step 2: HCl + NaOH → NaCl + H₂O is one to one, so moles of base equal moles of acid. Swap in sulfuric acid and that free ride vanishes — H₂SO₄ needs two NaOH per molecule, so a factor of ½ has to be inserted between steps 1 and 2 by hand, and the chain no longer works as written.

The classic wreck is dividing by the burette volume in step 2. The moles came out of the burette; they were matched by acid that came out of the pipette, so it is the 25.00 mL aliquot that fixes the unknown concentration. Steps 3 and 4 then scale the result to the full 250.0 mL and convert to a mass, which is how the answer gets used in practice. Volumetric analysis in this recognisable shape — burette, pipette, clamp stand, colour-change indicator — was codified by Karl Friedrich Mohr in his 1855 titrimetry textbook, and a Grade 11 bench would still look familiar to him.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.