In: Chemistry
Exercise 18.69
Exercise 18.69
Consider the titration of a 22.0 −mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. Determine each of the following.
Part A
the initial pH
Express your answer using two decimal places.
pH = | 2.85 |
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Significant Figures Feedback: Your answer 2.86 was either rounded differently or used a different number of significant figures than required for this part.
Part B
the volume of added base required to reach the equivalence point
V = | 18.6 | mL |
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Correct
Part C
the pH at 4.00 mL of added base
Express your answer using two decimal places.
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pH = |
4.43 |
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Part D
the pH at one-half of the equivalence point
Express your answer using two decimal places.
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pH = |
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Part E
the pH at the equivalence point
Express your answer using two decimal places.
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pH = |
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First of all, volumes will change in every case (Volume of base + Volume of acid), therefore the molarity will change as well
M = moles / volume
pH = -log[H+]
a) HA -> H+ + A-
Ka = [H+][A-]/[HA]
a) no NaOH
Ka = [H+][A-]/[HA]
Assume [H+] = [A-] = x
[HA] = M – x
Substitute in Ka
Ka = [H+][A-]/[HA]
Ka = x*x/(M-x)
1.8*10^-5 = x*x/(0.11-x)
This is quadratic equation
x =0.001398
For pH
pH = -log(H+)
pH =-log(0.001398)
pH in a = 2.85
b)
Volume required
mmol of acid = mmol of base
Macid*Vacid = Mbase*Vbase
Vbase = Macid*Vacid/Mbase
Vbase = 0.11*22/0.13 = 18.62 mL
Vbase = 18.62 mL required
c) 4 ml Base
This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)
Use Henderson-Hassebach equations!
pH = pKa + log (A-/HA)
initially
mmol of acid = MV = 0.11*22= 2.42 mmol of acid
mmol of base = MV = 0.13*4= 0.52 mmol of base
then, they neutralize and form conjugate base:
mmol of acid left = 2.42 -0.52 = 1.9 mmol
mmol of conjguate left = 0 + 0.52 = 0.52
Get pKa
pKa = -log(Ka)
pKa = -log(1.8*10^-5) = 4.75
Apply equation
pH = pKa + log ([A-]/[HA]) =
pH = 4.75+ log (0.52/1.95) = 4.18
d) for half equivalence
This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)
Use Henderson-Hassebach equations!
NOTE that in this case [A-] = [HA]; Expect pH lower than pKa
pH = pKa + log (A-/HA)
pH = pKa + log(1)
pH = pKa
pH = 4.75
e) Addition of Same quantitie of Acid/Base
This will be Hydrolisis (equilibrium of acid-base) and the weak acid/base will form an equilibrium
We will need Kb
Ka*Kb = Kw
And Kw = 10^-14 always at 25°C for water so:
Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.55*10^-10
Now, proceed to calculate the equilibrium
H2O + A- <-> OH- + HA
Then K equilibrium is given as:
Kb = [HA][OH-]/[A-]
Assume [HA] = [OH-] = x
[A-] = M – x
Substitute in Kb
5.55*10^-10 = [x^2]/[M-x]
recalculate M:
mmol of conjugate = 2.42 mmol
Total V = V1+V2 = 22+18.6 = 40.6
[M] = 2.42 mmol/40.6= 0.0596 M
5.55*10^-10 = [x^2]/[0.0596-x]
x = 5.75*10^-6
[OH-] = 5.75*10^-6
Get pOH
pOH = -log(OH-)
pOH = -log ( 5.75*10^-6) = 5.24
pH = 14-pOH = 14-5.24= 8.76
e) Addition of base
There will be finally an Excess of Base!
mol of acid < mol of base
Calculate pOH directly
[OH-] = M*V / Vt
mmol of acid = MV = 200*0.10 = 20
mmol of base = MV = 250*0.1 = 25
therefre,
mmol of strng base left = 25-20 = 5 mmol
Vtotal = 250+200 = 450 mL
[OH-] = 5/450 = 0.01111
pOH = -log(OH-)
pOH = -log(0.01111) = 1.954
pH = 14-pOH = 14-1.954= 12.046
pH = 12.046