Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.200 M HClO(aq) with 0.200 M KOH(aq). The ionization constant for HClO can be found asfunction _parent.followURL,http://sites.google.com/site/chempendix/ionization

(a) before addition of any KOH

(b) after addition of 25.0 mL of KOH

(c) after addition of 30.0 mL of KOH

(d) after addition of 50.0 mL of KOH

(e) after addition of 60.0 mL of KOH

Answer 1

millimoles of HClO= 50 x 0.2= 10

a) 0 ml KOH added

pH = 1/2 (pKa- log C)

   = 1/2 (7.4 -log (0.200) )

pH= 4.05

(b) after addition of 25.0 mL of KOH

millimoles of KOH = 0.20 x 25 = 5

this is half - equivalence point .so here

pH = pKa

pH = 7.40

(c) after addition of 30.0 mL of KOH

millimoles of KOH = 0.2 x 30 = 6

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

10           6                                      0               0 -----------------------initial

4         0                                           6      6     -----------------equilibirum

pH = pKa + log[salt/acid]

    = 7.4 + log (6/4)

    = 7.58

pH = 7.58

(d) after addition of 50.0 mL of KOH

millimoles of KOH = 0.2 x 50 = 10

this is equivalence point. so salt only remiains.

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

[salt] = salt millimoles /total volume in ml

           = 10/(50+50)

           = 0.1 M

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

   = 7 + 1/2 [7.4 + log (0.1)]

    = 10.2

pH = 10.2

e) after addition of 60.0 mL of KOH

millimoles of KOH = 0.2 x 60 = 12

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

10           12                                      0               0 -----------------------initial

0          2                                          10            10-------------------equilibirum

in the solution strong base remained

[base ] = 2/total volume = 2/110 = 0.018

pOH = -log[OH-] = -log(0.018) = 1.74

pH + pOH = 14

pH = 12.26