A solution containing 75.0 mL of 0.150 M strong acid (HCl) is titrated with 75.0 mL of 0.300 M strong base (NaOH). What is the pH of the resulting solution

12.875

Explanation:

Balanced equation of the reaction:

HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)

mole of acid = mole of base

75 * 0.150 = volume of base needed * 0.3

volume of base needed = 37.5 ml

excess OH = 0.3 * 37.5 ml / 150 ml where total volume = 75 ml + 75 ml = 150 ml

excess OH = 0.075 M

pOH = - log (OH⁻) = - log (0.075 M) = 1.125

pH + pOH = 14

pH = 14 - pOH = 14 - 1.125 = 12.875