A titration between a 0.1 M strong acid and 0.1 M strong base is performed in the lab. The student stops the titration when 15 mL of the base is added to 25 mL of acid in the flask.

Question

Determine the pH of the titration at this point.

1.78

13

3

1

Answer 1

Strong acid = H⁺

Strong base = OH^-

H⁺ + OH^- ==> H2O

moles H⁺ = 25 ml x 1 L /1000 ml x 0.1 mol/L = 0.0025 mols H⁺

moles OH^- = 15 ml x 1 L / 1000 ml x 0.1mol/L = 0.0015 mols OH^-

Excess H⁺ = 0.0025 mol - 0.0015 mol = 0.001 mols H⁺

Total volume = 40 ml = 0.04 L

[H⁺] = 0.001 mol H⁺ / 0.04 L = 0.025 M

pH = -log [H⁺]

pH = -log 0.025

pH = 1.6

None of the choices are correct as far as I can tell