An aqueous solution contains 0.162 M hydrosulfuric acid and 0.171 M hydroiodic a

Question

An aqueous solution contains 0.162 M hydrosulfuric acid and 0.171 M hydroiodic acid.

Calculate the sulfide ion concentration in this solution.
[S2-] =  mol/L.

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 solution contains two acids: 0.162M Hydrosulfuric acid (H2S) and 0.171M Hydroiodicacid (HI).

Hydrosulfuric acid (H2S), being diprotic acid has two steps of dissociation:

H2S -----> HS- + H+ This being the first step of dissociation,

Ka1 = {[HS-][H+]}/[H2S] --------- 1

HS- -----> S2- + H+ This being the second step of dissociation,

Ka2 = {[S2-][H+]}/[HS-] --------- 2

We also have Hydroiodic acid, HI in the solution. Since HI is a stong acid, it would completely dissociate in the following form:

HI --------> H+ + I-   ----------- 3

Due to common ion effect, dissociatoion of H2S is depressed and the only H+ ion concentration in solution is from HI.

Now going back to equation 1:

Ka1 = 1.0 X 10-7 (from the table)

[H+] = 0.171 M same as concentration of HI

[H2S] = 0.162 M, given.

Rearranging equation 1 and substituting these values, we can calulate [HS-]

[HS-] = Ka1[H2S]/[H+] = (1.0 X 10-7)(0.162)/(0.171) = 0.95 X 10-7 M

Rearranging equation 2 and substituting for [HS-] and Ka2 we can calulate [S2-]

[S2-] = Ka2[HS-]/[H+] = (1.0 X 10-19)(0.95 X 10-7)/(0.171) = 5.5 X 10-26 M

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