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Calculate the ph of a 0.100 m solution of acetic acid. ka = 1.8 x 10-5

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(H⁺)²= Ka x c(CH₃COOH)= 0,000018mol/dm³ x 0,1mol/dm³
(H⁺)²= 0,0000018 mol²/dm⁶
(H⁺)=√0,0000018= 0,00134mol/dm³
pH= -log(H⁺)=2,87
User Steven De Salas
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