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

1. Moles of hydroxide added: 0.100 M x 0.03750 L = 0.003750 mol. 2. Moles of hydronium present: 0.100 M x 0.02500 L = 0.002500 mol. 3. Hydroxide is now in excess by 0.003750 minus 0.002500 = 0.001250 mol. 4. Total volume is 25.00 plus 37.50 = 62.50 mL, which is 0.06250 L. 5. Concentration: 0.001250 divided by 0.06250 = 0.0200 mol per litre. 6. Take the negative logarithm: pOH = 1.70. 7. Convert: pH = 14.00 minus 1.70 = 12.30. 8. Reporting 1.70 is the trap of quoting the pOH as if it were the pH. _Source: OpenStax Chemistry (CC BY 4.0), Ch 14 "Acid-Base Equilibria", section 14.7 Acid-Base Titrations_