What is the approximate ph of a 0.06 m solution of ch3cooh (given that ka = 1.78 x 10-5)?

The equilibrium reaction of CH3COOH is CH3COOH = CH3COO(-) + H(+). When x moles of CH3COOH dissolves from an initial concentration of 0.06M, the equilibrium species will be [CH3COOH] = 0.06 - x; [CH3COOH-]=x; and [H+]=x. Now using the equilibrum expression for the acid you have: Ka = [CH3COOH-] [H+] / [CH3COOH] = x^2 / (0.06 - x). For a low Ka, as 1.78 * 10^ -5 is, 0.06 << x => 0.06 - x = 0.06. So, you can solve for x from 1.78 * 10^ -5 = x^2 / 0.06 => x^2 = 0.06 * 1.78 * 10^-5 = 1.068 * 10^ 6 => x = 1.033 * 10^ -3. So, now you can calculate the pH from pH = log { 1 / [H+] } = pH = log { 1 / (1.033*10^-3) } = 2.99. Answer pH = 2.99

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