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.

(I) To calculate the concentration of HS- ion
The case I (in the absence of HCI):
Let the concentration of HS- be x M.
         H2S     H+   +   HS-Ci      0.1            0              0Cf      0.1-x      x              xThen, Kai=H+HS-HsS9.1×10-8=xx0.1-x9.1×10-80.1-x=x2
Taking 0.1 - x M; 0.1 M, we have 9.1×10-80.1=x2.
9.1×10-9=x2x=9.1×10-9   =9.54×10-5MHS-=9.54×10-5M
Case ll (in the presence of HCl):
In the presence of 0.1 M of HCl, let HS- be y M
Then,        H2S     HS-   +   H+Ci      0.1            0              0Cf      0.1-y      y              yNow,  Then, Kai=HS-H+HsS9.1×10-8=y×0.10.1          0.1-y ; 0.1 M                                              and 0.1+y ; 0.1 M             9.1×10-8=yHS-=9.1×10-8
To calculate the concentration of S2-
Case l (in the absence of 0.1 M HCl):
HS-     H++S2- HS-=9.54×10-5M      From first ionization, case ILet  S2- be X.Also, H+=9.54×10-5M        From first ionization, case IKa2=[H+][S2][HS]Ka2=9.54×105X9.54×1051.2×1013=X=S2
Case ll (in the presence of 0.1 M HCl):
Again, let the concentration of HS- be X'M.
[S2]=XThen,Ka2=[H+][S2][HS]1.2×1013=(0.1)(X)9.1×10810.92×1021=0.1X10.92×10210.1=XX=1.092×10200.1=1.092×1019MKai=1.74×105