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A student has a solution of acetic acid (CH3COOH) with a concentration of 0.1 M and wants to calculate the effect of adding sodium acetate (NaCH3COO) with a concentration of 0.05 M on the equilibrium position of the reaction CH3COOH + H2O ↔ CH3COO- + H3O+. Calculate the new equilibrium concentrations of all species and explain the effect of the added common ion on the equilibrium position.

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To calculate the new equilibrium concentrations of all species, we can use the concept of the common ion effect and the equilibrium constant  Ka  for acetic acid. The Ka value for acetic acid is 1.8 x 10^-5.First, let's write the equilibrium expression for the reaction:Ka = [CH3COO-][H3O+] / [CH3COOH]Initially, the concentrations are as follows:[CH3COOH] = 0.1 M[CH3COO-] = 0.05 M  from the added sodium acetate [H3O+] = 0  assuming no dissociation has occurred yet Let x be the change in concentration of the species as the reaction reaches equilibrium:[CH3COOH] = 0.1 - x[CH3COO-] = 0.05 + x[H3O+] = xNow, substitute these expressions into the Ka equation:1.8 x 10^-5 =  0.05 + x  x  /  0.1 - x Since Ka is very small, we can assume that x is also very small compared to the initial concentrations. Therefore, we can simplify the equation by assuming that 0.05 + x  0.05 and 0.1 - x  0.1:1.8 x 10^-5 =  0.05  x  /  0.1 Now, solve for x:x =  1.8 x 10^-5  0.1  / 0.05x = 3.6 x 10^-6Now we can find the new equilibrium concentrations:[CH3COOH] = 0.1 - x  0.1 - 3.6 x 10^-6  0.1 M[CH3COO-] = 0.05 + x  0.05 + 3.6 x 10^-6  0.050004 M[H3O+] = x  3.6 x 10^-6 MThe addition of the common ion  CH3COO-  from sodium acetate causes the equilibrium to shift to the left, according to Le Chatelier's principle. This means that the dissociation of acetic acid  CH3COOH  is suppressed, resulting in a lower concentration of H3O+ ions and a higher concentration of CH3COO- ions at equilibrium. This is the common ion effect, which causes the pH of the solution to increase  become less acidic  due to the decrease in H3O+ concentration.

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