Exercise 18.69 Exercise 18.69 Consider the titration of a 22.0 mL sample of 0.11

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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.

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Correct

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

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Correct

Part C

the pH at 4.00 mL of added base

Express your answer using two decimal places.

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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Part E

the pH at the equivalence point

Express your answer using two decimal places.

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Explanation / Answer

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

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