The first ionization constant of H2S is 9.1×10−8. Calculate the concentration of HS− ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also?
If the second dissociation constant of H2S is 1.2×10−13, calculate the concentration of S2– under both conditions.
The first ionization constant of H2S is 9.1×10−8. Calculate the concentration of HS− ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also?
If the second dissociation constant of H2S is 1.2×10−13, calculate the concentration of S2– under both conditions.
(i) To calculate the concentration of HS– ion:
Case I (in the absence of HCl):
Let the concentration of HS– be x M.
H2S↔H++HS−C,0.100C,0.1−xxx
Then, Ka1=[H+][HS−][H2S]9.1×10−8=(x)(x)0.1−x(9.1×10−8)(0.1−x)=x2
Taking 0.1−xM;0.1M,wehave(9.1×10−8)(0.1)=x29.1×10−9=x2x=√9.1×10−9=9.54×10−5M⇒[HS−]=9.54×10−5M
Case II (in the presence of HCl):
In the presence of 0.1 M of HCl, let [HS−] be y M
Then,H2S↔HS−+H+C,0.100C,0.1−yyy
Now, Ka1=[HS−][H+][H2]Ka1=[y](0.1+y)(0.1−y)9.1×10−8=y×0.10.1 (∵0.1−y;0.1M)9.1×10−8=y (and 0.1+y;0.1M)9.1×10−8=y⇒[HS−]=9.1×10−8
To calculate the concentration of [S2−]
Case I (in the absence of 0.1 M HCl):
HS−↔H+S2−
[HS−]=9.54×10−5M (From first ionization Case I)
Let [S2−] be x
Also, [H+]=9.54×10−5M (From first ionization, case I)
Ka2=[H+][S2−][HS−]Ka2=(9.54×10−5)(X)9.54×10−51.2×10−13=X=[S2−]
Case II (in the presence of 0.1 M HCl):
Again, let the concentration of HS^– be X' M.
[HS−]=9.1×10−8M (From first ionization, case II)
[H+]=0.1M (From HCl, case II)
[S2−]=X′
Then, Ka2=[H+][S2−]HS−
1.2×10−13=(0.1)(X′)9.1×10−810.92×10−21=0.1X′10.92×10−210.1=X′X′=1.092×10−200.1=1.092×10−19M⇒Ka1=1.74×10−5
(i) To calculate the concentration of HS– ion:
Case I (in the absence of HCl):
Let the concentration of HS– be x M.
H2S↔H++HS−C,0.100C,0.1−xxx
Then, Ka1=[H+][HS−][H2S]9.1×10−8=(x)(x)0.1−x(9.1×10−8)(0.1−x)=x2
Taking 0.1−xM;0.1M,wehave(9.1×10−8)(0.1)=x29.1×10−9=x2x=√9.1×10−9=9.54×10−5M⇒[HS−]=9.54×10−5M
Case II (in the presence of HCl):
In the presence of 0.1 M of HCl, let [HS−] be y M
Then,H2S↔HS−+H+C,0.100C,0.1−yyy
Now, Ka1=[HS−][H+][H2]Ka1=[y](0.1+y)(0.1−y)9.1×10−8=y×0.10.1 (∵0.1−y;0.1M)9.1×10−8=y (and 0.1+y;0.1M)9.1×10−8=y⇒[HS−]=9.1×10−8
To calculate the concentration of [S2−]
Case I (in the absence of 0.1 M HCl):
HS−↔H+S2−
[HS−]=9.54×10−5M (From first ionization Case I)
Let [S2−] be x
Also, [H+]=9.54×10−5M (From first ionization, case I)
Ka2=[H+][S2−][HS−]Ka2=(9.54×10−5)(X)9.54×10−51.2×10−13=X=[S2−]
Case II (in the presence of 0.1 M HCl):
Again, let the concentration of HS^– be X' M.
[HS−]=9.1×10−8M (From first ionization, case II)
[H+]=0.1M (From HCl, case II)
[S2−]=X′
Then, Ka2=[H+][S2−]HS−
1.2×10−13=(0.1)(X′)9.1×10−810.92×10−21=0.1X′10.92×10−210.1=X′X′=1.092×10−200.1=1.092×10−19M⇒Ka1=1.74×10−5
