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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Win,odd Dhamnekar on December 20, 2020, 01:24:43 AM
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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 S2-under both conditions.
Solution: Concentration of S2- under both the conditions, are different in these two answers. Which one is correct? In my opinion, concentration of S2- in answer 1 is correct.
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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-], [S2-] (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.
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You are right that answer 1 is correct. The fallacy in answer 2 is the statement that [H+] = 2[S2-]. The correct mass balance is
[H+] = [HS-] + 2[S2-]
As the degree of the second ionisation is very small, we can use our answers for [HS-] and [H+] from the first part.
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Here is the neat, legible copy of first answer in which the concentration of [HS]- and S2- 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.
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This assumption does not take into account that the main amount of H+ comes from the 1st degree of dissociation of H2S
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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