7.45. 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.

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Vishal Baghel | Contributor-Level 10

5 months ago

(i) To calculate [HS⁻] in absence of HCl:
Let, [HS⁻] = x M.

H_2?S? H⁺ + HS⁻
The initial concentrations of H_2?S, H⁺ and HS⁻ are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-x M, x M and x M respectively.

K_a? = [H₂? S] [H⁺] [HS⁻]?
9.1×10⁻⁸ = x × x / (0.1−x)?
In the denominator, 0.1-x can be approximated to 0.1 as x is very small.
9.1×10⁻⁸=x×x / (0.1)?

x=9.54×10⁻⁵M= [HS⁻]

(ii) To calculate [HS⁻] in presence of HCl:
Let [HS⁻]= y M.

H₂? S? H

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(i) To calculate  [HS−] in absence of HCl: Let,   [HS−] = x M.H2? S? H+ + HS− The initial concentrations of H2? S, H+ and HS− are 0.1 M, 0 M and 0 M respectively. Their final concentrations are 0.1-x M, x M and x M respectively.Ka? = [H2? S] [H+] [HS−]?   9.1×10−8 = x × x / (0.1−x)? In the denominator, 0.1-x can be approximated to 0.1 as x is very small.    9.1×10−8=x×x / (0.1)? x=9.54×10−5M= [HS−] (ii) To calculate  [HS−] in presence of HCl: Let  [HS−]= y M. H2? S? H++HS− The initial concentrations of H2? S, H+ and HS− are 0.1 M, 0 M and 0 M respectively. Their final concentrations are 0.1-y M, y M and y M respectively. Also, HCl? H++Cl− For HCl  [H+]= [Cl−]=0.1M Ka? = [HS−] [H+] / [H2? S]? Ka? = y (0.1+y)? / (0.1−y) In the denominator, 0.1-y can be approximated to 0.1 and in the numerator, 0.1+y can be approximated to 0.1 as y is very small. 9.1×10−8= (y×0.1) / 0.1? y= 9.1×10−8M= [HS−] (iii) To calculate [S2−] in absence of 0.1 M HCl:HS−? H++S2− [HS−]=9.54×10−5M Let  [S2−]= x M. [H+]= 9.54×10−5 MKa? = [H+] [S2−]? / [HS−] = >1.2×10−13= (9.54×10−5×X? ) / (9.54×10−5) = > x=1.2×10−13M= [S2−] (iv) To calculate  [S2−] in presence of 0.1 M HCl: Let  [S2−]= x′ M. [HS−]= 9.1×10−8 M [H+]= 0.1 M (from HCl)Ka? = [HS−] [H+] [S2−]? 1.2×10−13= (0.1×X′? ) / (9.1×10−8)x′= 1.092×10−19= [S2−]

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