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calculate the ph of 1.00 l of a buffer that is 1.00 m in acetic acid and 1.00 m in sodium acetate after the addition of 0.400 mole of naoh.

Respuesta :

Initial moles of sodium acetate = acetic acid = 1 m * 1 L = 1 moles

Concentration of acetate after adding NaOH = 1 mole + 0.40 mole = 1.40 M

Concentration of acetic acid after adding NaOH = 1 mol - 0.40 mol = 0.60 M

According to Handerson - Hasselblach equation:

pH = Pka + ㏒ [ acetate]/[acetic acid]

we know that,

for acetic acid, pKa = 4.74

so, by substitution:

∴pH = 4.74 + ㏒[1.40/0.60]

∴ pH = 5.10

To know more about pH:

brainly.com/question/15289741

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