Question

The next three (3) problems deal with the titration of 471 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11) with 1.9 M NaOH.

1) What is the pH of the solution at the 2nd equivalence point?

2) What will the pH of the solution be when 0.1577 L of 1.9 M NaOH are added to the 471 mL of 0.501 M carbonic acid?

3) How many mL of the 1.9 M NaOH are needed to raise the pH of the carbonic acid solution to a pH of 5.842?

Answer 1

ans)

a)

What is the pH of the solution at the 2nd equivalence point?

At 2nd equivalence point H2CO3 is completely reacted with KOH and forms K2CO3 and the net reaction is shown below

H₂CO₃ + 2OH^- <===> CO₃^2- + H₂O

moles of H2CO3 taken = Molarity x volume = 0.471 x 0.501 = 0.236 mol H2CO3

Moles of KOH reacted with moles of H2CO3

0.236 mol H2CO3*2 mol KOH/1mol  H2CO3=

=0.472 mol KOH

Volume of KOH required =

0.472 mol KOH*1L/1.9

0.248 L

Total volume of the solution at this point = 0.471 + 0.248 = 0.719 L

Molarity of carbonate ions = moles/volume = 0.236/0.719 = 0.33 M

	CO32-	H2O <===>	HCO3-	OH-
Initial	0.33 M	-	0	0
change	-x	-	+x	+x
Equilibrium	0.33-x	-	x	x

Given Ka2 = 5.6 x 10^-11

Kb2 = Kw/Ka2 =

Kb2=1.786 x 10^-4

Kb2=x^2/0.33-x

neglecting denominator x

1.786*10^-4(0.33)=x^2

x =0.0077 M

[OH-] = 0.0077 M

pOH = -log[OH-]

pOH = 2.11

pH = 14-2.11 = 11.89 = 11.9

pH = 11.9

b)

What will the pH of the solution be when 0.1577 L of 1.9 M KOH are added to the 471 mL of 0.501 M carbonic acid?

Moles of KOH = 1.9 x 0.1577 = 0.2996 mol

Moles of H2CO3 = 0.501 x 0.471 =0.236 mol

Initially H2CO3 is completely converted to HCO3-

H2CO3 + OH- ---> HCO3- + H2O

The excess KOH further reacts with HCO3- to give carbonate ion

moles of KOH left after reacting wiith H2CO3 = 0.2996-0.236 = 0.0636 mol

HCO₃^- + OH^- <===> CO₃^2- + H₂O

Initial moles of HCO3- = 0.236 mol

Moles of HCO3 after reacting with 0.0636 mol KOH = 0.236-0.0636 = 0.172 mol HCO3-

Moles of CO32- formed = moles of KOH = 0.0636 mol

Using Henderson -Hasselbalch equation

pH=-log(5.68*10^-11)+log [ 0.0636]/[0.172]

pH = 10.25 -0.43

pH = 9.82

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c)

How many mL of the 1.9 M KOH are needed to raise the pH of the carbonic acid solution to a pH of 5.842?

pKa1 = -log(4.3 x 10^-7)

pKa1 = 6.37

pH < pKa1

So we have to consider pH

H₂CO₃ <==> HCO₃^- + H⁺

Let x moles of KOH is added

H₂CO₃ + OH^- ---> HCO₃^- + H₂O

	H2CO3	OH- --->	HCO3-	H2O
Initial	0.236 mol	x	0	-
Change	-x	-x	+x	-
after	0.236-x	0	x

Henderson Hasselbalch eqaution

pH=pka1+log[HCO3-]/[H2CO3)

5.842=6.37+log[x]/[0.236-x]

-0.528=log[x]/[0.236-x]

[x]/[0.236-x]=10^-0.528=0.296

x = 0.296(0.236 - x)

x = 0.0698 - 0.296 x

1.296x = 0.0698

x = 0.054 mol

Moles of KOH = x = 0.054 mol

Given Molarity of KOH = 1.9 M

Volume of KOH = 0.054/1.9 = 0.0284 L = 28.4 mL