Calculate the pH for each case in the titration of 50.0 mL of 0.160 M HClO(aq) with 0.160 M KOH(aq).
ionization constant=4.0×10^–8
1)when 0.0 mL of KOH is added
HClO dissociates as:
HClO -----> H+ + ClO-
0.16 0 0
0.16-x x x
Ka = [H+][ClO-]/[HClO]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((4*10^-8)*0.16) = 8*10^-5
since c is much greater than x, our assumption is correct
so, x = 8*10^-5 M
use:
pH = -log [H+]
= -log (8*10^-5)
= 4.0969
Answer: 4.10
2)when 25.0 mL of KOH is added
Given:
M(HClO) = 0.16 M
V(HClO) = 50 mL
M(KOH) = 0.16 M
V(KOH) = 25 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.16 M * 50 mL = 8 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.16 M * 25 mL = 4 mmol
We have:
mol(HClO) = 8 mmol
mol(KOH) = 4 mmol
4 mmol of both will react
excess HClO remaining = 4 mmol
Volume of Solution = 50 + 25 = 75 mL
[HClO] = 4 mmol/75 mL = 0.0533M
[ClO-] = 4/75 = 0.0533M
They form acidic buffer
acid is HClO
conjugate base is ClO-
Ka = 4*10^-8
pKa = - log (Ka)
= - log(4*10^-8)
= 7.398
use:
pH = pKa + log {[conjugate base]/[acid]}
= 7.398+ log {5.333*10^-2/5.333*10^-2}
= 7.398
Answer: 7.40
3)when 40.0 mL of KOH is added
Given:
M(HClO) = 0.16 M
V(HClO) = 50 mL
M(KOH) = 0.16 M
V(KOH) = 40 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.16 M * 50 mL = 8 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.16 M * 40 mL = 6.4 mmol
We have:
mol(HClO) = 8 mmol
mol(KOH) = 6.4 mmol
6.4 mmol of both will react
excess HClO remaining = 1.6 mmol
Volume of Solution = 50 + 40 = 90 mL
[HClO] = 1.6 mmol/90 mL = 0.0178M
[ClO-] = 6.4/90 = 0.0711M
They form acidic buffer
acid is HClO
conjugate base is ClO-
Ka = 4*10^-8
pKa = - log (Ka)
= - log(4*10^-8)
= 7.398
use:
pH = pKa + log {[conjugate base]/[acid]}
= 7.398+ log {7.111*10^-2/1.778*10^-2}
= 8
Answer: 8.00
4)when 50.0 mL of KOH is added
Given:
M(HClO) = 0.16 M
V(HClO) = 50 mL
M(KOH) = 0.16 M
V(KOH) = 50 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.16 M * 50 mL = 8 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.16 M * 50 mL = 8 mmol
We have:
mol(HClO) = 8 mmol
mol(KOH) = 8 mmol
8 mmol of both will react to form ClO- and H2O
ClO- here is strong base
ClO- formed = 8 mmol
Volume of Solution = 50 + 50 = 100 mL
Kb of ClO- = Kw/Ka = 1*10^-14/4*10^-8 = 2.5*10^-7
concentration ofClO-,c = 8 mmol/100 mL = 0.08M
ClO- dissociates as
ClO- + H2O -----> HClO + OH-
0.08 0 0
0.08-x x x
Kb = [HClO][OH-]/[ClO-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((2.5*10^-7)*8*10^-2) = 1.414*10^-4
since c is much greater than x, our assumption is correct
so, x = 1.414*10^-4 M
[OH-] = x = 1.414*10^-4 M
use:
pOH = -log [OH-]
= -log (1.414*10^-4)
= 3.8495
use:
PH = 14 - pOH
= 14 - 3.8495
= 10.1505
Answer: 10.15
5)when 60.0 mL of KOH is added
Given:
M(HClO) = 0.16 M
V(HClO) = 50 mL
M(KOH) = 0.16 M
V(KOH) = 60 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.16 M * 50 mL = 8 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.16 M * 60 mL = 9.6 mmol
We have:
mol(HClO) = 8 mmol
mol(KOH) = 9.6 mmol
8 mmol of both will react
excess KOH remaining = 1.6 mmol
Volume of Solution = 50 + 60 = 110 mL
[OH-] = 1.6 mmol/110 mL = 0.0145 M
use:
pOH = -log [OH-]
= -log (1.455*10^-2)
= 1.8373
use:
PH = 14 - pOH
= 14 - 1.8373
= 12.1627
Answer: 12.16
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