[Win,odd Dhamnekar]

The-first ionization constant of H₂S is 9.1 x 10^-8. Calculate the concentration of HS^- ions in its 0.1 M solution and how will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H₂S is 1.2 x 10^-13, calculate the concentration of S^2-under both conditions.


Solution:  Concentration of S^2- under both the conditions, are different in these two answers. Which one is correct? In my opinion, concentration of S^2- in answer 1 is correct.

[AWK]

Copies of solutions are not very legible.
You ignore significant figures. For concentrations 0.1 - 1 significant digit in the answer, for 0.10 - two significant digits.

The correct approximate solutions are as follows [HS^-], [S^2-] (for water and acid, respectively)
 - SQRT (0.1·K₁), K₂, K₁ and K₁K₂/0.1, respectively).
The calculator is only useful to calculate the square root and the product of K₁·K₂.

What does the last line in Part Three mean.

[mjc123]

You are right that answer 1 is correct. The fallacy in answer 2 is the statement that [H⁺] = 2[S^2-]. The correct mass balance is
[H⁺] = [HS^-] + 2[S^2-]
As the degree of the second ionisation is very small, we can use our answers for [HS^-] and [H⁺] from the first part.

[Win,odd Dhamnekar]

Here is the neat, legible copy of first answer in which the concentration of [HS]^- and S^2- in the absence of HCl and in the presence of HCl is computed correctly. I also don't  understand last line of part three of second answer.

[AWK]

This assumption does not take into account that the main amount of H⁺ comes from the 1st degree of dissociation of H₂S

[Win,odd Dhamnekar]

Neat and legible second answer to this question.
Solution:-
 Given: First ionization constant of H₂S= 9.1×10^-8.
 The  second dissociation constant of H₂S is 1.2 × 10^-13.
(i) To compute the concentration of HS^- ion:
Case a: in the absense of HCl: concentration of H₂S solution =0.1M