Question

Calculate the pH for each of the following cases in the titration of 50.0mL of 0.160M HClO(aq) with 0.160M KOH(aq).

a) before addition of any KOH

b) after addition of 25.0 mL of KOH

c) after addition of 40.0 mL of KOH

d) after addition of 50.0 mL of KOH

e) after addition of 60.0 mL of KOH

Answer 1

pKa = -logKa = = -log(4.0 x10^ -8.) = 7.4

millimoles of HClO= 50 x0.160= 8

a) 0 ml KOH added

pH = 1/2 (pKa- log C)

   = 1/2 (7.4 -log (0.160) ) = 4.10

pH= 4.10

(b) after addition of 25.0 mL of KOH

it is first equivalece point here pH = pKa

pH = 7.4

(c) after addition of 40.0 mL of KOH

millimoles of KOH = 40 x 0.160 = 6.4

HClO + KOH ------------------------------> KClO + H2O

8            6.4 0               0 -----------------------initial

1.6 0                                           6.4 6.4-------------------equilibirum

in the solution acid and salt remained so it can form buffer

For acidic buffer

Henderson-Hasselbalch equation

pH = pKa + log[salt/acid]

    = 7.4 + log (6.4/1.6)

    = 8.00

pH = 8.00

(d) after addition of 50.0 mL of KOH

millimoles of KOH = 0.160 x 50 = 8

HClO + KOH ------------------------------> KClO + H2O

8        8 0               0 -----------------------initial

0          0 8 8----------------equilibirum

in the solution salt remained so we have to use salt hydrolysis.

it is the salt of strong base and weak acid so pH should be more than 7

[salt] = salt millimoles /total volume in ml

           = 8 /(50+50)

           = 0.08 M

pH = 7 + 1/2[Pka + logC]

   = 7 + 1/2 [7.4 + log (0.08)]

    = 10.15

pH = 10.15

e) after addition of 60.0 mL of KOH

millimoles of KOH = 0.160 x 60 = 9.6

HClO + KOH ------------------------------> KClO + H2O

8            9.6 0               0 -----------------------initial

0          1.6 8 8-----------------equilibirum

in the solution strong base remained

[base ] = 1.6 /total volume = 1.6 /110 = 0.0145

pOH = -log[OH-] = -log(0.0145) =1.84

pH + pOH = 14

pH = 12.16