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What is the approximate pH of a .06 M solution of CH3COOH given that Ka= 1.78*10-5

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CH₃COOH ⇔ CH₃COO⁻ + H⁺

[CH₃COO⁻] = [H⁺] = x


K=(|CH_(3)COO^(-)|*|H^(+)|)/(|CH_(3)COOH|)=(x^(2))/(|CH_(3)COOH|)\\\\ 1,78*10^(-5)=(x^(2))/(0,06) \ \ |*0,06\\\\ 0,1068*10^(-5)=x^(2)\\\\ x_(1)\approx0,001 \ \land \ \ x_(2)=\approx-0,001\\\\ pH=-log|H^(+)|=-log0,001=3
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