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What is the pH of a 0.1 M solution of acetic acid (CH3COOH) if its dissociation constant (Ka) 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 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 represent the change in concentration as follows:CH3COOH: 0.1 - xCH3COO-: xH+: xNow, we can substitute these values into the Ka expression:1.8 x 10^-5 =  x  x / 0.1 - x Assuming that x is much smaller than 0.1, we can simplify the equation: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 find the pH using the definition of pH:pH = -log10[H+]pH = -log10 1.34 x 10^-3 pH  2.87Therefore, the pH of the 0.1 M solution of acetic acid is approximately 2.87.
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