When a 27.2 mL sample of a 0.371 M aqueous hydrocyanic acid solution is titrated

Question

When a 27.2 mL sample of a 0.371 M aqueous hydrocyanic acid solution is titrated with a 0.316 M aqueous potassium hydroxide solution, what is the pH at the midpoint in the titration?

pH =

note

Acid/Base Ionization Constants at 25 oC

Acid

Formula

Ka1

Ka2

Ka3

Acid/Base Ionization Constants at 25 oC

Acid

Formula

Ka1

Ka2

Ka3

Acetic acid

CH3COOH

1.8×10-5

Acetylsalicylic acid (aspirin)

HC9H7O4

3.0×10-4

Aluminum ion

Al(H2O)43+

1.2×10-5

Arsenic acid

H3AsO4

2.5×10-4

5.6×10-8

3.0×10-13

Ascorbic acid

H2C6H6O6

7.9×10-5

1.6×10-12

Benzoic acid

C6H5COOH

6.3×10-5

Carbonic acid

H2CO3

4.2×10-7

4.8×10-11

Ferric ion

Fe(H2O)63+

4.0×10-3

Formic acid

HCOOH

1.8×10-4

Hydrocyanic acid

HCN

4.0×10-10

Hydrofluoric acid

HF

7.2×10-4

Hydrogen peroxide

H2O2

2.4×10-12

Hydrosulfuric acid

H2S

1.0×10-7

1.0×10-19

Hypochlorous acid

HClO

3.5×10-8

Nitrous acid

HNO2

4.5×10-4

Oxalic acid

H2C2O4

5.9×10-2

6.4×10-5

Phenol

C6H5OH

1.0×10-10

Phosphoric acid

H3PO4

7.5×10-3

6.2×10-8

3.6×10-13

Sulfuric acid

H2SO4

very large

1.2×10-2

Sulfurous acid

H2SO3

1.7×10-2

6.4×10-8

Zinc ion

Zn(H2O)42+

2.5×10-10

Base

Formula

Kb

Ammonia

NH3

1.8×10-5

Aniline

C6H5NH2

7.4×10-10

Caffeine

C8H10N4O2

4.1×10-4

Codeine

C18H21O3N

8.9×10-7

Diethylamine

(C2H5)2NH

6.9×10-4

Dimethylamine

(CH3)2NH

5.9×10-4

Ethylamine

C2H5NH2

4.3×10-4

Hydroxylamine

NH2OH

9.1×10-9

Isoquinoline

C9H7N

2.5×10-9

Methylamine

CH3NH2

4.2×10-4

Morphine

C17H19O3N

7.4×10-7

Piperidine

C5H11N

1.3×10-3

Pyridine

C5H5N

1.5×10-9

Quinoline

C9H7N

6.3×10-10

Triethanolamine

C6H15O3N

5.8×10-7

Triethylamine

(C2H5)3N

5.2×10-4

Trimethylamine

(CH3)3N

6.3×10-5

Urea

N2H4CO

1.5×10-14

Acid/Base Ionization Constants at 25 oC

Acid

Formula

Ka1

Ka2

Ka3

Acid/Base Ionization Constants at 25 oC

Acid

Formula

Ka1

Ka2

Ka3

Acetic acid

CH3COOH

1.8×10-5

Acetylsalicylic acid (aspirin)

HC9H7O4

3.0×10-4

Aluminum ion

Al(H2O)43+

1.2×10-5

Arsenic acid

H3AsO4

2.5×10-4

5.6×10-8

3.0×10-13

Ascorbic acid

H2C6H6O6

7.9×10-5

1.6×10-12

Benzoic acid

C6H5COOH

6.3×10-5

Carbonic acid

H2CO3

4.2×10-7

4.8×10-11

Ferric ion

Fe(H2O)63+

4.0×10-3

Formic acid

HCOOH

1.8×10-4

Hydrocyanic acid

HCN

4.0×10-10

Hydrofluoric acid

HF

7.2×10-4

Hydrogen peroxide

H2O2

2.4×10-12

Hydrosulfuric acid

H2S

1.0×10-7

1.0×10-19

Hypochlorous acid

HClO

3.5×10-8

Nitrous acid

HNO2

4.5×10-4

Oxalic acid

H2C2O4

5.9×10-2

6.4×10-5

Phenol

C6H5OH

1.0×10-10

Phosphoric acid

H3PO4

7.5×10-3

6.2×10-8

3.6×10-13

Sulfuric acid

H2SO4

very large

1.2×10-2

Sulfurous acid

H2SO3

1.7×10-2

6.4×10-8

Zinc ion

Zn(H2O)42+

2.5×10-10

Explanation / Answer

The reaction is

HCN + KOH -----------------> KCN + H2O

27.2x0.371=10.09 0 0 0 initial mmoles

5.045 0 5.045 - at mid point of titration

Thus the solution is a buffer whose pH is given by Hendersen equation as

pH = pKa + log [KCN]/ [HCN]

Ka of HCN = 4.0x10-10

and pKa = 9.3979

Thus pH = 9.3979 + log 5.045/5.045

= 9.3979

For a weak acid -strng base titration pH = pKa at the mid point or half equivalence point.

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