Solubility of a metal (M) sulphide, MS in water is
10-10. The pH of the medium needed to prevent
precipitation of this sulphide using 0.1 M HS
(Ka, *Ka, of H2S = 10-20) from 0.1 M solution of
M2+ is
(1) 1
(2) 2
(3) 3
(4) 4
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Answered by
27
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Solubility of a metal (M) sulphide, MS in water is
Solubility of a metal (M) sulphide, MS in water is10-10. The pH of the medium needed to prevent
Solubility of a metal (M) sulphide, MS in water is10-10. The pH of the medium needed to preventprecipitation of this sulphide using 0.1 M HS
Solubility of a metal (M) sulphide, MS in water is10-10. The pH of the medium needed to preventprecipitation of this sulphide using 0.1 M HS(Ka, *Ka, of H2S = 10-20) from 0.1 M solution of
Solubility of a metal (M) sulphide, MS in water is10-10. The pH of the medium needed to preventprecipitation of this sulphide using 0.1 M HS(Ka, *Ka, of H2S = 10-20) from 0.1 M solution ofM2+ is
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3 is the correct option....
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Answered by
1
The correct option is (2) 2.
- To prevent the precipitation of the metal sulfide MS from the solution, we need to maintain the concentration of the sulfide ion HS- below the solubility product of MS.
- This can be achieved by adding an acid (H+) to the solution, which will react with the HS- ion to form more H2S.
- This will shift the equilibrium to the left, reducing the concentration of HS- ion and preventing the precipitation of MS.
Using the equation pH = pKa + log ([HS-]/[H2S])
Ka = [H+][HS-]/[H2S]
10^-20 = [H+][HS-]/0.1
[H+][HS-] = 10^-22
[HS-] = (10^-22/[H+])
Substituting the value of [HS-] into the pH equation :
pH = 6.3 + log (10^-12)
pH = 2
Therefore, the pH of the solution needed to prevent the precipitation of MS using 0.1 M HS from a 0.1 M solution of M2+ is 2.
To learn more about pH from the given link.
https://brainly.in/question/31227
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