Hydrogen sulfide (H2S) is an odorous gas that forms the following equilibrium wh

Question

Hydrogen sulfide (H2S) is an odorous gas that forms the following equilibrium when dissolved in water:

H2S(aq) H + (aq) + HS– (aq)

H2S can be removed from water by a process called stripping. The water is contacted with steam, air, or another gas. The hydrogen sulfide is then absorbed into the gas as long as it is in the H2S form.

a) Find the fraction of hydrogen sulfide in the H2S form at a pH of 5 and 9.

b) Should the H2S stripping process be operated at high or low pH?

c) At what pH will the fraction be 0.50 = 50%?

Answers:

a) At a pH of 5, 0.992 = 99.2%

At a pH of 9, 0.012 =1.2%

a) low pH.

b) pH of 7.1.

Explanation / Answer

Ka of H2S = 9.1*10^-8
let the initial concentration of H2S be 1 M
at pH = 5

[H+] from acid =10^-5

[H+] from water = 10^-6

total [H+] = 1.1*10^-5
                 H2S <--> H+   +   HS-
initial :             1           1.1*10^-5 0
at equilibrium: 1-x      1.1*10^-5+x       x

Ka = [H+] [HS-]/ [H2S]
9.1*10^-8 = (1.1*10^-5+x)*x / (1-x)
x will be very small, so it can be ignored in terms of 1
9.1*10^-8 = (1.1*10^-5+x)*x / 1
x^2 + 1.1*10^-5 x - 9.1*10^-8 = 0
solve for x

x=2.96*10^-4

percent of H2S form = (1-x)*100 /1 =(1-x)*100 = 99.97%

at pH = 9

[H+] from acid= 10^-9

[H+] from water = 10^-8

total [H+] = 1.1*10^-8
                 H2S <--> H+   +   HS-
initial :             1           1.1*10^-8 0
at equilibrium: 1-x      1.1*10^-8+x       x

Ka = [H+] [HS-]/ [H2S]
9.1*10^-8 = (1.1*10^-8+x)*x / (1-x)
x will be very small, so it can be ignored in terms of 1
9.1*10^-8 = (1.1*10^-8+x)*x / 1
x^2 + 1.1*10^-8 x - 9.1*10^-8 = 0
solve for x

3.01*10^-4

percent of H2S form = (1-x)*100 /1 = (1-x)*100 = 99.96%

b)

At high pH, [H+] will be low and [H2S] will be low (Le chattelier's principle).
so, it should be done at low pH

c)

let pH

                 H2S <--> H+   +   HS-
initial :             1           10^-pH      0
at equilibrium: 1-x       10^-pH+x     x

x = 0.5

9.1*10^-8 = (10^-pH+0.5)*(0.5)

10^-pH+0.5 = 1.82*10^-7

10^-pH = negative value

so, it is not possible

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