Consider the titration of 40.0 mL of 0.171-M of KX with 0.141-M HCl. The pKa of HX = 10.89. 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 24.7 mL of acid? pH = e) What is the pH of the solution after the addition of 58.2 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