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What is the expected pH of a 0.1 M solution of acetic acid (CH3COOH) with a pKa value of 4.76? Provide a detailed explanation of your answer, including any necessary calculations or formulas.

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To calculate the expected pH of a 0.1 M solution of acetic acid, we will use the Henderson-Hasselbalch equation, which relates the pH, pKa, and the ratio of the concentrations of the conjugate base  A-  and the weak acid  HA  in the solution. The equation is as follows:pH = pKa + log  [A-]/[HA] First, we need to determine the concentrations of the acetic acid  HA  and its conjugate base, the acetate ion  A- . Since acetic acid is a weak acid, it will only partially dissociate in water:CH3COOH  HA   CH3COO-  A-  + H+Initially, we have 0.1 M of acetic acid and 0 M of acetate ion. Let x be the concentration of acetic acid that dissociates. At equilibrium, we will have  0.1 - x  M of acetic acid and x M of acetate ion.Now, we can write the expression for the acid dissociation constant  Ka  of acetic acid:Ka = [CH3COO-][H+]/[CH3COOH] = [x][x]/[0.1 - x]Given the pKa value of 4.76, we can calculate the Ka value:Ka = 10^-pKa  = 10^-4.76   1.74 x 10^-5 Now, we can substitute the Ka value into the equation:1.74 x 10^-5  = [x][x]/[0.1 - x]Since Ka is very small, we can assume that x is much smaller than 0.1, so we can simplify the equation:1.74 x 10^-5   x^2/0.1Now, we can solve for x:x^2  1.74 x 10^-5  * 0.1x^2  1.74 x 10^-6 x  1.32 x 10^-3 Now that we have the concentration of the acetate ion  A-  and the remaining concentration of acetic acid  HA , we can use the Henderson-Hasselbalch equation:pH = pKa + log  [A-]/[HA] pH = 4.76 + log  1.32 x 10^-3  /  0.1 - 1.32 x 10^-3   pH  4.76 + log  1.32 x 10^-3  / 0.1 pH  4.76 - 1pH  3.76Therefore, the expected pH of a 0.1 M solution of acetic acid with a pKa value of 4.76 is approximately 3.76.

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