pH = 12.69
Explanation
concentration acetic acid = 0.9400 M
volume acetic acid = 54.4 mL
moles acetic acid = (concentration acetic acid) * (volume acetic acid)
moles acetic acid = (0.9400 M) * (54.4 mL)
moles acetic acid = 51.136 mmol
moles NaOH added = (concentration NaOH) * (volume NaOH)
moles NaOH added = (0.4300 M) * (141 mL)
moles NaOH added = 60.63 mmol
excess moles OH^{-} = (moles NaOH added) - (moles acetic acid)
excess moles OH^{-} = (60.63 mmol) - (51.136 mmol)
excess moles OH^{-} = 9.494 mmol
[OH^{-}] = (excess moles OH^{-}) / (total volume)
[OH^{-}] = (9.494 mmol) / (54.4 mL + 141 mL)
[OH^{-}] = 0.0486 M
pOH = -log[OH^{-}]
pOH = -log(0.0486 M)
pOH = 1.31
pH = 14 - pOH
pH = 14 - 1.31
pH = 12.69
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