Consider the titration of a 24.0 mL sample of 0.100 M HC2H3O2 with 0.120 M NaOH.

Question

Consider the titration of a 24.0 mL sample of 0.100 M HC2H3O2 with 0.120 M NaOH. Determine each of the following.

Part A

the initial pH

Express your answer using two decimal places.

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Part B

the volume of added base required to reach the equivalence point

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Part C

the pH at 6.00 mL of added base

Express your answer using two decimal places.

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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.

Consider the titration of a 24.0 mL sample of 0.100 M HC2H3O2 with 0.120 M NaOH. Determine each of the following.

Part A

the initial pH

Express your answer using two decimal places.

pH =

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Part B

the volume of added base required to reach the equivalence point

V =   mL  

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Part C

the pH at 6.00 mL of added base

Express your answer using two decimal places.

pH =

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Part D

the pH at one-half of the equivalence point

Express your answer using two decimal places.

pH =

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Part E

the pH at the equivalence point

Express your answer using two decimal places.

pH =

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/(0.1-x)

This is quadratic equation

x =0.001332

For pH

pH = -log(H+)

pH =-log(0.001332)

pH in a = 2.87

b)

V = mmol of NaOH/ M of NaOh

mmol of NaOH = mmol of acid

mmol of acid = MV = 24*0.1 = 2.4

V = 2.4/0.120

V = 20 mL of base

c) 6 ml NaOH

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)

initially

mmol of acid = MV =2.4

mmol of base = MV = 6*0.12= 0.72 mmol of base

then, they neutralize and form conjugate base:

mmol of acid left = 2.4-0.72   = 1.68

mmol of conjguate left = 0 + 0.72   = 0.72   5

Get pKa

pKa = -log(Ka)

pKa = -log(1.8*10^-4) = 4.75

Apply equation

pH = pKa + log ([A-]/[HA]) =

pH = 4.75+ log (0.72/1.68 ) = 4.382

d) for half equiv ml

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)

from pKa = 4.75

pH = 4.75 + log(1)

pH = 4.75

d) 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 = 20 mmol

Total V = V1+V2 = 24+20= 44

[M] = 2.4/44= 0.0545 M

5.55*10^-10 = [x^2]/[0.0545 -x]

x =0.00000547

[OH-] =0.00000547

Get pOH

pOH = -log(OH-)

pOH = -log (0.00000547) = 5.26

pH = 14-pOH = 14-5.26= 8.74

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