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

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asked Jul 8, 2018 112k views

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Anaphory · Answer 1 · 2018-07-13T11:11:23+0000

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

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