Question

An analytical chemist is titrating 54.4 ml of a 0.9400 M solution of acetic acid (HCH,CO,)... An analytical chemist is titrating 54.4 ml of a 0.9400 M solution of acetic acid (HCH,CO,) with a 0.4300 M solution of NaOH. The pk of acetic acid is 4.70. Calculate the pit of the acid solution after the chemist has added 141 ml of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places, pH = 0 xs ?

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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