To evaluate the pH, we need to do a hydrolysis equilibrium problem. calculate the equilibrium constant, Kb, for this reaction from the fact that the KaKb = Kw for any acid/conjugate base pair (or in other instances, for any base/conjugate acid pair). Set up a table to determine the equilibrium concentrations in solution. Initially, 0.067 M benzoate is present in solution, to establish equilibrium an amount of benzoate must be “x” used to form an amount of benzoic acid and amount of hydroxide. Combining “x” “x” these quantities gives us the equilibrium quantities: + H2O(l) C6H5CO2H(aq) + initial 0.067 M 0 0 change -x + x + x 0.067 x x x – Substitute these quantities into the equilibrium constant expression: The in the denominator term, 0.067 x can be neglected because the magnitude of the “x” – constant is 107 smaller than the last certain digit of the known concentration, hence the expression will simplify to: for x: x2 = (1.6 x 10-10)(0.067) = 1.1 x 10-11 There are two common methods for finding the pH. One method would be to find the pOH and then convert to pH: pOH = = -log[3.3 x 10-6] = 5.48 pH = 14 pOH = 14 5.48 = 8.52 – – Alternatively, one could first calculate the [H3O+] using the Kw expression for water and then find the pH: pH = -log[H3O+] = -log(3.04 x 10-9) = 8.52
To evaluate the pH, we need to do a hydrolysis equilibrium problem. calculate the equilibrium constant, Kb, for this reaction from the fact that the KaKb = Kw for any acid/conjugate base pair (or in other instances, for any base/conjugate acid pair). Set up a table to determine the equilibrium concentrations in solution. Initially, 0.067 M benzoate is present in solution, to establish equilibrium an “x” amount of benzoate must be used to form an “x” amount of benzoic acid and “x” amount of hydroxide. Combining these quantities gives us the equilibrium quantities: + H2O(l) C6H5CO2H(aq) + initial 0.067 M 0 0 change -x + x + x 0.067 – x x x Substitute these quantities into the equilibrium constant expression: The “x” in the denominator term, 0.067 – x can be neglected because the magnitude of the constant is 107 smaller than the last certain digit of the known concentration, hence the expression will simplify to: for x: x2 = (1.6 x 10-10)(0.067) = 1.1 x 10-11 There are two common methods for finding the pH. One method would be to find the pOH and then convert to pH: pOH = = -log[3.3 x 10-6] = 5.48 pH = 14 – pOH = 14 – 5.48 = 8.52 Alternatively, one could first calculate the [H3O+] using the Kw expression for water and then find the pH: pH = -log[H3O+] = -log(3.04 x 10-9) = 8.52