Question

Assignment Score: 3450/5000 Resources 9 Hint Check Answer Question 8 of11 The pK, of hypochlorous acid is 7.530. A 57.0 ml. solution of 0.118 M sodium hypochlorite (NaOCI) is titrated with 0.335 M HCI Calculate the pH of the solution after the addition of 7.41 ml. of 0.335 M HCİ Calculate the pli of the solution after the addition of 20.9 ml. of 0.335 M HC the phi of the solution after the addition of 20.9 ml. of 0.335 M H Calculate the plH of the solution at the equivalence point with 0335 M HCI BANG 144 8 5

Answer 1

a)

millimoles of NaOCl = 57 x 0.118 = 6.726

millimoles of HCl = 7.41 x 0.335 = 2.482

NaOCl     +    HCl   -----------------------> HClO   + NaCl

6.726              2.482                                    0              0

4.244              0.00                                  2.482          ---

pH = pKa + log [NaOCl / HClO ]

pH = 7.530 + log (4.244 / 2.482)

pH = 7.76

b)

millimoles of HCl = 20.9 x 0.335 = 7.0015

[H+]left = 7.0015 -6.726 / (20.9 + 57)

            = 3.54 x 10^-3 M

pH = -log [H+]

pH = 2.45

c)

equivalence point volume = 20.08 mL

[HOCl] formed = 6.726 / ( 57 + 20.08) = 0.0873 M

pH = 7 + 1/2 [pKa +log C]

pH = 7 + 1/2 [7.530 + log 0.0873]

pH = 10.24