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  • The first ionization constant of H_2S is 9.1 × 10^–8. Calculate the concentration of HS^– ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also ?

7.45   The first ionization constant of H2S is 9.1 × 10-8. Calculate the concentration of HS- ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H2S is 1.2 × 10-13, calculate the concentration of S2- under both conditions.

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best_answer

We have,
1st ionisation constant of hydrogen sulphide is 9.1\times 10^{-8} and the 2nd dissociation constant is 1.2\times 10^{-13}

Case 1st-(absence of hydrochloric acid)

To calculate the concentration of HS^-
Let x be the concentration of HS^-and the ionisation of hydrogen sulphide is;
H_2S\rightleftharpoons H^++HS^-
 0.1 M

At equilibrium, the concentration of various species are,
Since the dissociation constant is is very small. So, x can be neglected.
the concentration of H_2S = 1-x M
the concentration of HS^- and H^+ is x M
So, 
K_a = \frac{x^2}{0.1}=9.1\times 10^{-8}
from here x can be calculated and we get,  x = 9.54\times 10^{-5}M

Case 2nd (In presence of 0.1 M, HCl)
Suppose H_2S is dissociated is x .Then at equilibrium, 
[H_2S] = 0.1-x\simeq 0.1,\ [H^+]=0.1+x\simeq 0.1 and the [HS^-] = x M
So, 
K_a = \frac{x(0.1)}{0.1}=9.1\times 10^{-8}
Thus the concentration of [HS^-] is 9.1\times 10^{-8}


 

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