12.50 mL of 0.100 M NaOH is added to 25.00 mL of 0.100 M HCl. What is the pH?

1. Moles of hydronium at the start: 0.100 M x 0.02500 L = 0.002500 mol. 2. Moles of hydroxide added: 0.100 M x 0.01250 L = 0.001250 mol. 3. These neutralise one for one, leaving 0.001250 mol of hydronium in excess. 4. Total volume is 25.00 plus 12.50 = 37.50 mL, which is 0.03750 L. 5. Concentration: 0.001250 divided by 0.03750 = 0.0333 mol per litre. 6. Take the negative logarithm: minus log of 0.0333 gives pH = 1.477. 7. Forgetting to allow for the extra volume would give a slightly lower pH. 8. The value 7.000 arrives only at 25.00 mL, when the two amounts match exactly. _Source: OpenStax Chemistry (CC BY 4.0), Ch 14 "Acid-Base Equilibria", section 14.7 Acid-Base Titrations_