Consider the titration of 35.0 mL of 0.165-M of KX with 0.131-M HCl. The pKa of HX = 7.91. Give all pH values to 0.01 pH units. a) What is the pH of the original solution before addition of any acid? pH = b) How many mL of acid are required to reach the equivalence point? VA = mL c) What is the pH at the equivalence point? pH = d) What is the pH of the solution after the addition of 30.0 mL of acid? pH = e) What is the pH of the solution after the addition of 52.9 mL of acid? pH =
Explanation / Answer
a)KX >> K+ + X- X- + H2O HX + OH- K = Kw/ Ka Ka = 10^-9.61 = 2.45 x 10^-10 K = 1.0 x 10^-14 / 2.45 x 10^-10 = 4.07 x 10^-5 4.07 x 10^-5 = x^2 / 0.223 - x x = [OH-] =0.00301 M pOH = - log 0.00301 =2.52 pH = 14 - 2.52 = 11.5 b) Moles X- = 0.0450 L x 0.233 = 0.0105 X- + H+ >> HX Moles H+ needed = 0.0105 V = moles / M = 0.0105 / 0.158 = 0.0664 L => 66.4 mL c) at eq. point total volume = 0.0664 L + 0.045 L= 0.111 L concentration HX = 0.0105 / 0.111 =0.0943 M HX H+ + X- 2.45 x 10^-10 = x^2 / 0.0943-x x = 4.81 x 10^-6 M => [H+] pH = 5.32