Question

1) Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH).

1 2 3 4 5  100.0 mL of 0.100 M KOH by 0.100 M HCl

1 2 3 4 5  200.0 mL of 0.100 M H₂NNH₂ (K_b = 3.0 x 10^-6) by 0.100 M HCl

1 2 3 4 5  100.0 mL of 0.100 M C₂H₅NH₂ (K_b = 5.6 x 10^-4) by 0.100 M HCl

1 2 3 4 5  100.0 mL of 0.100 M HCl by 0.100 M NaOH

1 2 3 4 5  100.0 mL of 0.100 M HF (K_a = 7.2 x 10^-4) by 0.100 M NaOH

Consider the major species present at the halfway point. From this, you should be able to deduce the order of the pH with only minimal calculations

Incorrect.

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2) Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH).

1 2 3 4 5  100.0 mL of 0.100 M HF (K_a = 7.2 x 10^-4) by 0.100 M NaOH

1 2 3 4 5  200.0 mL of 0.100 M H₂NNH₂ (K_b = 3.0 x 10^-6) by 0.100 M HCl

1 2 3 4 5  200.0 mL of 0.100 M HC₂H₃O₂ (K_a = 1.8 x 10^-5) by 0.100 M NaOH

1 2 3 4 5  100.0 mL of 0.100 M KOH by 0.100 M HCl

1 2 3 4 5  100.0 mL of 0.100 M C₂H₅NH₂ (K_b = 5.6 x 10^-4) by 0.100 M HCl

Answer 1

1.

A.

mmoles of KOH = 100*0.100 = 10

At half equivalence point

mmoles of HCl = 50*0.100= 5.

mmoles of excess KOH = (10 - 5)= 5.

Molarity = ( moles/ volume)

= (5/50+100)

= 0.033 M.

pOH = - log [OH^-] = - log (0.033) = 1.477

Or, pH = 14 -1.47 = 12.53 .

B.

Half equivalence point , [ NH₂NH₂] = [NH₂NH₃⁺]

pOH = pKb = - logKb = - log(3.0*10^-6) = 5.522

Or, pH = 14 -5.52 = 8.48.

C.

Half equivalence point , [C₆H₅NH₃⁺] = [C₆H₅NH₂]

pOH = pKb = - log Kb = - log (5.6*10^-4) = 3.25.

pH = 14 - 3.25 = 10.75 .

D.

mmoles of HCl = (100*0.100) = 10.

At half equivalence point

mmoles of NaOH = (50*0.100)= 5.

Excess mmoles of HCl = (10-5) = 5

Molarity of HCl = 5/(50+100) = 0.033 M.

pH = - log[H⁺] = - log(0.033) = 1.48

E.

At half equivalence point [HF] = [NaF]

pH = pKa = - log Ka = - log(7.2*10^-4) = 3.14.

Hence rank is

D = 1

E = 2

B = 3

C = 4

A = 5

2.

At equivalence point

100 ml 0.1 M HF = 100 ml 0.1 NaOH

mmoles NaF = 100*0.1 = 10.

[NaF] = 10/(100+100) = 0.05 M

pH is due to dissociation of the salt

Hence, pH = 7 + ( pKa + logC)

= 7 + ( 3.14 + log 0.05)

= 7+0.92 = 7.92.

B.

[ NH₂NH₃⁺] = (200*0.100/(200+200) = 0.05 M.

pH of salt of weak base strong acid.

Hence, pH = 7 - ( pK_b+ logC)

= 7 -   ( 5.522 - log 0.05)

= 7 - (5.522 -1.30)

= 7 - 2.1 = 4.9.

C.

[C₂H₃O₂] = [200*0.100/(200+200) ] = 0.05 M.

Hence , pH of salt solution of weak acid strong base

pH = 7 + (pKa + logC)

=7 + ( - log1.8*10^-5 + log 0.05)

= 7+1.725

= 8.725.

D.

pH at equivalence point if strong acid (HCl) and KOH = 7.

E.

[C₂H₅NH₃⁺] =[( 100*0.100)/100+100) ] = 0.05 M.

Then

pH = 7 -   ( pKb + logC)

= 7 -   ( -log5.6*10^-4 + log0.05)

= 7 -   (3.25 -1.30)

= 6.025.

Hence ranking is

B = 1

E = 2

D = 3

A = 4

C = 5