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

idk i’m struggling too

c. is the answer

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