2 years ago

What is the ph of a 0.135 m aqueous solution of potassium acetate, kch3coo? (ka for ch3cooh = 1.8×10-5)?

1 answer:

6 0

[OH-] = √(Cs×Kw)/Ka
[OH-] = √(0,135×10^-14)/1,8×10^-5
[OH-] = √0,075×10^-9 = √75×10^-12 = 8,66×10^-6

pOH = -log[OH-]
pOH = -log[8,66×10^-6]
pOH ≈ 5,0625

pH + pOH = 14
pH = 14-5,0625
pH = 8,9375

:•)

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