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Titration to molarity and mass of solute

SCH3U Grade 11 Chemistry · Solutions and Solubility

During a stockroom inventory, a technician finds a 250.0 mL volumetric flask of clear liquid whose label has peeled away; the requisition book says dilute hydrochloric acid, but the concentration must be established before the flask can go back on the shelf. A 25.00 mL aliquot is pipetted into a clean Erlenmeyer flask and two drops of phenolphthalein are added, while a burette on the clamp stand is rinsed and filled with standardised 0.1000 mol/L sodium hydroxide. The base is run in with constant swirling, fast at first, then drop by drop as each pink flash takes longer to fade. The endpoint — the first faint pink that persists in the swirling flask — 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.

0.1000 mol/L NaOHtitre 22.50 mL25.00 mL aliquotwhite tile

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • V_flask = 250 mL — Volumetric flask of unknown acid
  • V_aliquot = 25 mL — Aliquot pipetted for the titration
  • C_NaOH = 0.1 mol/L — Standardised sodium hydroxide
  • V_titre = 22.5 mL — Titre at the phenolphthalein endpoint
  • M = 36.46 g/mol — Molar mass of HCl
Determine
  1. (a)the moles of base the burette delivered
  2. (b)the concentration of the acid
  3. (c)the moles of HCl in the whole flask
  4. (d)the mass of HCl the flask holds
Step 1 of 4(a) · 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.

VCn
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=2.25 mmoln = 2.25\ \text{mmol}

Carried onward at full precision, not this rounded figure.

Open the Molarity (C = n/V) solver →

Step 2 of 4(b) · 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.

VCn
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=90 mol/m3C = 90\ \text{mol/m}^{3}

Carried onward at full precision, not this rounded figure.

Open the Molarity (C = n/V) solver →

Step 3 of 4(c) · solve for Amount of solute

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

VCn
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=22.5 mmoln = 22.5\ \text{mmol}

Carried onward at full precision, not this rounded figure.

Open the Molarity (C = n/V) solver →

Step 4 of 4(d) · solve for Mass

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

mMn
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=820.35 mgm = 820.35\ \text{mg}

Carried onward at full precision, not this rounded figure.

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

Answer

Therefore the burette delivered 2.250 mmol of base, the acid stands at 0.0900 mol/L, the full flask holds 22.50 mmol of HCl, and the label can finally read: dilute hydrochloric acid, 0.820 g in 250.0 mL.

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.