Question

. A solution of hypochlorous acid (pKa = 7.53) with an initial volume of 50.0 mL and a formal concentration of 0.120 M is titrated with 0.120 M potassium hydroxide. Complete the table below by calculating the pH of the hypochlorous acid solution after the specified volumes of titrant are added. Then, circle the principle species. If both species are present at an equal concentration, circle both. Show all work on separate sheet(s). Volume of titrant added (mL) pH Principle Species (circle) 0.00 HOCl OCl– 25.0 HOCl OCl– 35.0 HOCl OCl– 50.0 HOCl OCl– 55.0 HOCl OCl–

Answer 1

1)molarity of HOCL initially =50*0.12/1000=0.006 M

HOCl + H₂O = H₃O⁺ + OCl^-

0.006 0 0

0.006-x x x

Ka =x²/0.006-x = 2.951

  x² + 2.951x-0.0177=0 x= 0.0059 pH =-log 0.0059=2.229(initial)

the major species are H⁺(aq) and OCl^-(aq) and they are equal

molarity of KOH = 0.120 M

2)molarity of HOCL after adding 25 ml HOCL = 75*0.12/1000=0.009M

Ka =x²/0.009-x = 2.951

  x² + 2.951x-0.0266=0,x= 0.0059

pH =-log 0.0089=2.050

3)molarity of HOCL after adding 35 ml HOCL =85*0.12/1000=0.0102M

Ka =x²/0.0102-x = 2.951

  x² + 2.951x-0.030=0 , x=0.01

pH =-log 0.01=2

4)molarity of HOCL after adding 50 ml HOCL =100*0.12/1000=0.012M

Ka =x²/0.012-x = 2.951

  x² + 2.951x-0.0354=0 ,x=0.012

pH =-log 0.012=1.92

5)molarity of HOCL after adding 55 ml HOCL =105*0.12/1000=0.0126M

Ka =x²/0.0126-x = 2.951

   x² + 2.951x-0.0371=0 ,x=0.0125

pH =-log 0.0125=1.90