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…
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Question “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…”
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).
Answer
1) When 0.0 mL KOH is added
HClO is a dissociating agent that can be used to:
HClO —–> + ClO-
0.16 0 0
0.16-xx x
Ka = [H+][ClO -]/[HClO]
Ka = x*x/c-x
Assuming that x is ignored can be compared to c
The above expression is now
Ka = x*x/(c).
so, x = sqrt (Ka*c)
x = sqrt = ((4*10-8*0.16) = (8*10-5).
Our assumption is correct because c is greater than x
So, x = 8*10-5M
use:
pH = -log [H+]
= -log (8*10^-5)
= 4.0969
Answer: 4.10
2) When 25.0 mL KOH is added
Given:
M(HClO = 0.16M
V(HClO), 50 mL
M(KOH) = 0.16 M
V(KOH) = 25 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO), = 0.16 M x 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 each will react
4 mmol excess HClO left
Volume of Solution = 50 + 25, = 75 mL
[HClO] = 4 mg/75 mL = 0.0533M
[ClO-] = 4/75 = 0.0533M
They create acidic buffer
Acid is HClO
ClO-conjugate base is the name of the conjugate base
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 KOH is added
Given:
M(HClO = 0.16M
V(HClO), 50 mL
M(KOH) = 0.16 M
V(KOH) = 40 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO), = 0.16 M x 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
Both will react at 6.4 mmol
Exceed HClO = 1.6 mmol
Volume of Solution = 50 +40 = 90 mL
[HClO] = 1.5 mmol/90mL = 0.02178M
[ClO-] = 6.4/90 = 0.0711M
They create acidic buffer
Acid is HClO
ClO-conjugate base is the name of the conjugate base
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 KOH is added
Given:
M(HClO = 0.16M
V(HClO), 50 mL
M(KOH) = 0.16 M
V(KOH) = 50 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO), = 0.16 M x 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 each will react to make ClO- or H2O
ClO- Here is a strong base
ClO- formed = 8 mmol
Volume of Solution = 50 +50 = 100 mL
Kb of ClO – Kw/Ka = 1*10-14/4*10-8 = 2.25*10-7
Concentration ofClO,c = 8 mg/100 mL = 0.08M
ClO- is a dissociated as
ClO- + H2O —–> CLO-OH-
0.08 0 0
0.08-xx x
Kb = [HClO][OH ClO]
Kb = x*x/c-x
Assuming that x is ignored can be compared to c
The above expression is now
Kb = x*x/(c).
so, x = sqrt (Kb*c)
x = sqrt ((2.5*10^-7)*8*10^-2) = 1.414*10^-4
Our assumption is correct because c is greater than x
So, x = 1.41*10-4M
[OH-] =x = 1.414*10-4M
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 KOH is added
Given:
M(HClO = 0.16M
V(HClO), 50 mL
M(KOH) = 0.16 M
V(KOH) = 60 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO), = 0.16 M x 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 each will react
Exceed KOH = 1.6 Mmol
Volume of Solution = 50 +60 = 110 mL
[OH-] = 1.5 mmol/110mL = 0.0445 M
use:
pOH = -log [OH-]
= -log (1.455*10^-2)
= 1.8373
use:
PH = 14. – pOH
= 14 – 1.8373
= 12.1627
Answer: 12.16
Conclusion
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