Question

Calculate the pH for each case in the titration of 50.0 mL of 0.180 M HClO(aq)0.180 M HClO(aq) with 0.180 M KOH(aq).0.180 M KOH(aq). Use the ionization constant for HClO.HClO.

What is the pH before addition of any KOH?

pH=pH=

What is the pH after addition of 25.0 mL KOH?

pH=

What is the pH after addition of 30.0 mL KOH?

pH=

What is the pH after addition of 50.0 mL KOH?

pH=

What is the pH after addition of 60.0 mL KOH?

pH=

Answer 1

millimoles of HClO= 50 x0.180= 9

a) 0 ml KOH added

pH = 1/2 (pKa- log C)

   = 1/2 (7.4 -log (0.180) ) = 4.10

pH= 4.07

(b) after addition of 25.0 mL of KOH

it is first equivalece point here pH = pKa

pH = 7.4

(c) after addition of 30.0 mL of KOH

millimoles of KOH = 30 x 0.180 = 5.4

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

9            5.4 0               0 -----------------------initial

3.6 0                                          5.4    5.4-------------------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 (5.4/3.6)

    = 7.58

pH = 7.58

(d) after addition of 50.0 mL of KOH

millimoles of KOH = 0.180 x 50 = 9

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

9        9    0               0 -----------------------initial

0          0    9    9 ----------------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

           = 9 /(50+50)

           = 0.09 M

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

   = 7 + 1/2 [7.4 + log (0.09)]

    = 10.18

pH = 10.18

e) after addition of 60.0 mL of KOH

millimoles of KOH = 0.180 x 60 = 10.8

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

9             10.8                                        0               0 -----------------------initial

0            1.8                                          9                 9----------------equilibirum

in the solution strong base remained

[base ] = 1.8 /total volume = 1.8 /110 = 0.0164

pOH = -log[OH-] = -log(0.0164) =1.79

pH + pOH = 14

pH = 12.21