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Calculate the pH of a 0.1 M solution of acetic acid (CH3COOH) given that its dissociation constant Ka is 1.8x10^-5.

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To calculate the pH of a 0.1 M solution of acetic acid, we can use the formula for the dissociation constant  Ka  and the definition of pH. The dissociation of acetic acid can be represented as:CH3COOH  CH3COO- + H+The Ka expression for this reaction is:Ka = [CH3COO-][H+]/[CH3COOH]Since the initial concentration of acetic acid is 0.1 M, we can assume that the change in concentration due to dissociation is small and can be represented by x. Thus, the concentrations at equilibrium are:[CH3COOH] = 0.1 - x[CH3COO-] = x[H+] = xNow, we can substitute these values into the Ka expression:1.8x10^-5 =  x  x / 0.1 - x Since x is small compared to 0.1, we can approximate  0.1 - x  as 0.1:1.8x10^-5 = x^2/0.1Now, solve for x:x^2 = 1.8x10^-5 * 0.1x^2 = 1.8x10^-6x = sqrt 1.8x10^-6 x  1.34x10^-3Since x represents the concentration of H+ ions, we can now calculate the pH:pH = -log[H+]pH = -log 1.34x10^-3 pH  2.87Therefore, the pH of a 0.1 M solution of acetic acid is approximately 2.87.
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