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What is the pH of a 0.1 M solution of acetic acid (CH3COOH) at 25°C, given that the Ka of acetic acid is 1.8 x 10^-5?

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To find the pH of a 0.1 M solution of acetic acid, we can use the Ka expression and the formula for pH. The Ka expression for acetic acid is:Ka = [H+][CH3COO-] / [CH3COOH]Since the initial concentration of acetic acid is 0.1 M, we can set up an equilibrium expression:1.8 x 10^-5 =  x  x  /  0.1 - x Here, x represents the concentration of H+ ions, which is equal to the concentration of CH3COO- ions at equilibrium. We can assume that x is much smaller than 0.1, so the expression becomes:1.8 x 10^-5 = x^2 / 0.1Now, we can solve for x:x^2 = 1.8 x 10^-5 * 0.1x^2 = 1.8 x 10^-6x =  1.8 x 10^-6 x  1.34 x 10^-3Since x represents the concentration of H+ ions, we can now calculate the pH using the formula:pH = -log[H+]pH = -log 1.34 x 10^-3 pH  2.87So, the pH of a 0.1 M solution of acetic acid at 25C is approximately 2.87.
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