Question

*** ONLY NEED C

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.250 M HClO(aq) with 0.250 M KOH(aq). The ionization constant for HClO can be found here.

(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 = 7.4

millimoles of HClO= 50 x0.250= 12.5

a) 0 ml KOH added

pH = 1/2 (pKa- log C)

   = 1/2 (7.4 -log (0.250) ) = 4.00

pH= 4.00

(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 0.250 = 10

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

12.5          10    0               0 -----------------------initial

2.5 0                                           10            -----------------equilibirum

pH = pKa + log[salt/acid]

    = 7.4 + log (10/2.5)

    = 8.00

pH = 8.00

(d) after addition of 50.0 mL of KOH

millimoles of KOH = 0.250 x 50 = 12.5

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]   = 12.5 /(50+50)

           = 0.125 M

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

   = 7 + 1/2 [7.4 + log (0.125)]

    = 10.25

pH = 10.25

e) after addition of 60.0 mL of KOH

millimoles of KOH = 0.250 x 60 = 15

pH = 12.36