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

[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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