pOH = - = - log(1.0 x 10-12) = 12.00 or pH + pOH = 14, so pOH = 14 pH = 14 2.00 = 12.00 – – 9 b. Ca(OH)2 is a strong base, thus it dissociates completely: Ca(OH)2(aq) Ca2+(aq) + 2 OH¯ → pH = - log[H3O+] = - log(1.7 x 10-14) = 13.78 pOH = - = - log(0.60) = 0.22 or pH + pOH = 14, so pOH = 14 pH = 14 13.78 = 0.22 – – 9 c. Benzoic acid, C6H5COOH, is a weak acid with a Ka = 6.2 x 10-5. Write the equilibrium reaction and determine the expressions for the equilibrium amounts, starting with an initial amount of 0.20 M C6H5COOH. We are assuming that initially, there isn’t any acetate or hydronium ion present. An x amount of benzoic acid has to be lost to form an x amount of both benzoate ion and hydronium ion. Thus, at equilibrium, an amount of 0.20 x of benzoic acid is present, and x amount of both benzoate ion and hydronium ion – are present. This can be conveniently summarized in the table shown below: C6H5COOH(aq) + H2O (l) + H3O+(aq) initial 0.20 M 0 0 change -x +x +x equil. 0.20 x x x – The equilibrium constant expression for the equilibrium is: Substituting the equilibrium amounts into the above expression gives:
pOH = - = - log(1.0 x 10-12) = 12.00 or pH + pOH = 14, so pOH = 14 – pH = 14 – 2.00 = 12.00 9 b. Ca(OH)2 is a strong base, thus it dissociates completely: Ca(OH)2(aq) → Ca2+(aq) + 2 pH = - log[H3O+] = - log(1.7 x 10-14) = 13.78 pOH = - = - log(0.60) = 0.22 or pH + pOH = 14, so pOH = 14 – pH = 14 – 13.78 = 0.22 9 c. Benzoic acid, C6H5COOH, is a weak acid with a Ka = 6.2 x 10-5. Write the equilibrium reaction and determine the expressions for the equilibrium amounts, starting with an initial amount of 0.20 M C6H5COOH. We are assuming that initially, there isn’t any acetate or hydronium ion present. An x amount of benzoic acid has to be lost to form an x amount of both benzoate ion and hydronium ion. Thus, at equilibrium, an amount of 0.20 – x of benzoic acid is present, and x amount of both benzoate ion and hydronium ion are present. This can be conveniently summarized in the table shown below: C6H5COOH(aq) + H2O (l) + H3O+(aq) initial 0.20 M 0 0 change -x +x +x equil. 0.20 – x x x The equilibrium constant expression for the equilibrium is: Substituting the equilibrium amounts into the above expression gives: OH¯
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