Question

4.) Consider the titration of 100.0 mL of 0.016 M HOCl (Ka=3.5×10−8) with 0.0400 M NaOH.

Part A

How many milliliters of 0.0400 M NaOH are required to reach the equivalence point?

Part B

Calculate the pH after the addition of 10.0 mL of 0.0400 M NaOH.

Part C

Calculate the pH halfway to the equivalence point .

Part D

Calculate the pH at the equivalence point.

Answer 1

milli moles of HOCl = 100 x 0.016 = 1.6

part A : we need 1.6 milli moles of NaOH

1.6 = V x 0.04

V = 1.6 / 0.04 40 mL

40.0 mL NaOH required to reach equivalent point.

part B:

when 10 mL 0.04 M NaOH added

milli moles of NaOH = 10 x 0.04 =0.4

milli moles of HOCl left = 1.6 - 0.4 = 1.2

0.4 milli moles of NaOCl formed.   so in this case   buffer will form with HOCl and NaOCl

pH of acidic buffer

pH = pKa + log (salt) / (acid)

pKa = - log (Ka

pKa = -log (3.5 x 10 ^-8)

pKa = 7.46

pH = 7.46 + log (0.4 / 110) / (1.2 / 110)

pH= 6.98

part C:

half way of equivalent point means 0.8 milli moles acid consumed 0.8 milli moles NaOCl formed 0.8 milli moles HOCl left

milli moles of NaOH = 0.8

0.8 = V x 0.04

V = 20

pH = 7.46 + log (0.8 / 120) / (0.8 / 120)

pH = 7.46

part D:

at equivalent point

1.6 milli moles of NaOCl will form

so we need 1.6 milli moles of NaOH

1.6 = V x 0.04

V = 1.6 / 0.04 = 40 mL

concentration of NaOCl = 1.6 / 140 = 0.0112 M

pH =1 / 2 ( pK_W + pK_a + log C)

pH = 1 / 2(14 + 7.46 + log 0.0112)

pH = 9.76