Given, 9.1×10-8 is the initial(first) ionization constant of the gas H2S.Find out the concentration of the ion HS– in a 0.1M solution of H2S . Find the changes in concentration if the concentration is 0.1M in HCl. Find the concentration of S2- under both conditions, if 1.2×10-13 is the second dissociation constant of H2S.
Given:
First ionization constant of H2S= 9.1×10-8
Second dissociation constant of H2S =1.2×10-13
[H2S] = 0.1M
[HCl] = 0.1M
To calculate the [HS–]
In the absence of 0.1M HCl
Let the concentration of HS- ion be xM
Then, Using the formula:
Ka = 
Ka1 = 
As Ka1 = 9.1 × 10–8 (given)
∴ 9.1×10-8 = 
⇒9.1×10-8 = 
⇒(9.1 × 10-8) (0.1 – x) = x2
Taking 0.1 – x 
0.1
⇒(9.1 × 10-8) (0.1) = x2
⇒x = √ 9.1 × 10-8
⇒x = 9.54 × 10-5 M
Thus, in the absence of HCl, the concentration of HS- ion is 9.54 × 10-5 M
In the presence of 0.1 M HCl, let the concentration of HS- ion be yM
Then at equilibrium
Then, using the formula:
Ka1 = 
Ka1 = 
As Ka1 = 9.1 × 10–8
∴ 9.1×10-8 = 
⇒(9.1 × 10-8)(0.1 - y) = 
Taking 0.1 – y 
0.1
0.1 + y≈ 0.1
∴ (9.1 × 10-8)(0.1) = y(0.1)
⇒y = 9.1 × 10-8 M
Thus, in the presence of 0.1M HCl, the concentration of HS- ion is 9.1 × 10-8 M
To calculate [S2–]
In the absence of HCl
Let the concentration of S2- ion be x M
HS-↔ H+ + S2–
[HS-] = 9.54 × 10-5M (from first ionization)
[H+] = 9.54 × 10-5M (from first ionization)
[S2-] = xM
By applying the formula:
Ka = 
Ka1 = 
As Ka2 = 1.2 × 10–13 (given)
∴ 1.2 × 10–13= 
⇒x = 1.2 × 10–13M
Thus, in the absence of HCl, the concentration of S2- ion is 1.2 × 10–13 M
Case 2: In the presence of 0.1 M HCl
Let the concentration of S2- ion be x M
HS-↔ H+ + S2–
[HS-] = 9.1 × 10-8 (from first ionization, Case 2)
[H+] = 0.1 M(from HCl, Case 2)
[S2-] = yM
By applying the formula:
Ka = 
Ka2 = 
As Ka2 = 1.2 × 10–13 (given)
∴ 1.2 × 10–13= 
⇒10.92 × 10-21 = 0.1 y
⇒y = 1.092 × 10–19M
Thus, in the presence of 0.1 M HCl, the concentration of S2- ion is 1.092 × 10–19 M