Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.160 M HClO(aq) with 0.160 M KOH(aq). The ionization constant for HClO can be found here....4.0*10-8

before addition of KOH?

After addition of 25.0 mL KOH?   Addition of 30.0 mL KOH?    50.0 mL KOH?    60.0 mL 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) ) = 3.37

pH= 4.10

(b) after addition of 25.0 mL of KOH

millimoles of KOH = 0.160 x 25 = 4

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

8         4                                              0               0 -----------------------initial

4         0                                             4             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 (4/4)

pH = 7.4

(c) after addition of 30.0 mL of KOH

millimoles of KOH = 0.160 x 30 = 4.8

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

8            4.8                                            0               0 -----------------------initial

3.2          0                                             4.8           4.8----------------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 (4.8 /3.2)

    = 7.58

pH = 7.58

(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 M

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

pH + pOH = 14

pH = 12.16