In: Chemistry
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.220 M HClO(aq) with 0.220 M KOH(aq). The ionization constant for HClO is 4.0e-8.
(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
pKa = -logKa = = -log(4.0 x10^ -8.) = 7.4
millimoles of HClO= 50 x0.220= 11
a) 0 ml KOH added
pH = 1/2 (pKa- log C)
= 1/2 (7.4 - log (0.220) ) = 4.03
pH= 4.03
(b) after addition of 25.0 mL of KOH
millimoles of KOH = 0.22 x 25 = 5.5
HClO + KOH ------------------------------> KClO + H2O
11 5.5 0 0 -----------------------initial
5.5 0 5.5 5.5 -------------------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 (5.5/5.5)
pH = 7.4
(c) after addition of 30.0 mL of KOH
millimoles of KOH = 0.22 x 30 = 6.6
HClO + KOH ------------------------------> KClO + H2O
11 6.6 0 0 -----------------------initial
4.4 0 6.6 6.6 -------------------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 (6.6/4.4)
= 7.57
pH = 7.57
(d) after addition of 50.0 mL of KOH
millimoles of KOH = 0.22 x 50 = 11
HClO + KOH ------------------------------> KClO + H2O
11 11 0 0 -----------------------initial
0 0 11 11-------------------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
= 11/(50+50)
= 0.11 M
pH = 7 + 1/2[Pka + logC]
= 7 + 1/2 [7.4 + log (0.11)]
= 10.22
pH = 10.22
e) after addition of 60.0 mL of KOH
millimoles of KOH = 0.22 x 60 = 13.2
HClO + KOH ------------------------------> KClO + H2O
11 13.2 0 0 -----------------------initial
0 2.2 11 11-------------------equilibirum
in the solution strong base remained
[base ] = 2.2/total volume = 2.2/110 = 0.02
pOH = -log[OH-] = -log(0.02) =1.7
pH + pOH = 14
pH = 12.3