Strong Acid–Strong Base Titration: HCl with NaOH

Given

• 25.00 mL of 0.100 M HCl (analyte)

• 0.100 M NaOH (titrant)

• Both are strong, so neutralization goes to completion: H+ + OH- -> H2O.

Find

(a) The volume of NaOH at the equivalence point, (b) the pH after 10.00 mL of NaOH has been added, (c) the pH at the equivalence point.

Principle

In a strong acid–strong base titration, each mole of OH- destroys one mole of H+. The equivalence point is reached when moles of titrant equal the initial moles of analyte. Before equivalence, the pH is determined by the excess strong acid divided by the total solution volume; at equivalence only the neutral salt remains, so pH = 7.

Formula

Solution

1. Compute initial moles of H+: n(HCl) = 0.02500 L x 0.100 mol/L = 0.00250 mol.

2. (a) At equivalence, mol NaOH = mol HCl = 0.00250 mol, so V(NaOH) = 0.00250 mol / 0.100 mol/L = 0.02500 L = 25.00 mL.

3. (b) After 10.00 mL NaOH: mol OH- added = 0.01000 L x 0.100 mol/L = 0.00100 mol.

4. Subtract: H+ remaining = 0.00250 - 0.00100 = 0.00150 mol. Total volume = 25.00 + 10.00 = 35.00 mL = 0.03500 L.

5. [H+] = 0.00150 mol / 0.03500 L = 0.0429 M, so pH = -log(0.0429) = 1.37.

6. (c) At equivalence all H+ and OH- have neutralized; the solution contains only NaCl in water, which does not hydrolyze, so pH = 7.00.

Unit check

Moles come from L x mol/L = mol, and volume from mol / (mol/L) = L, so every division is dimensionally consistent. Concentrations are mol/L throughout, and pH is dimensionless.

What this means physically

This is the classic strong-strong titration: the equivalence point lands exactly at pH 7 because the salt formed (NaCl) is neutral — neither Na+ nor Cl- hydrolyzes. Before equivalence the pH is set by the leftover strong acid, and the sharp jump through pH 7 is what makes indicators like phenolphthalein work here.