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.

Asked by Pragya Singh | 1 year ago |  111

##### Solution :-

To calculate [HS]

To find [HS]

Case 1 – HCl is absent.

Now

Ka = $$\dfrac{ ([H^{+}][HS^{–}])}{[H_2S] }$$

= 9.1×10-8 (given)

$$\dfrac{ x^2}{(0.1-x)}$$

= 9.1×10-8

But 0.1-x is approximately equal to 0.1. Substituting this value in the equation:

$$\dfrac{x^2}{0.1}$$ = 9.1×10-8

x2 = 9.1×10-9

x = 9.54× 10-5 M

concentration of HS is 9.54×10-5 M.

Case 2 – HCl is present

Now,

Ka = $$\dfrac{(H+HS^{–})}{H_2S}$$

= (y× (0.1+y))/(0.1-y)

= 9.1×10-8 (given)

But (0.1 + y) and (0.1 – y) can be approximated to 0.1.

9.1× 10-8 = $$\dfrac{(0.1\times y)}{0.1}$$

y = [HS] = 9.1× 10-8 M

To calculate [S2-]:

Case 1 – HCl is absent.

The dissociation of HS is given by the equation:

HS ⇌ H+ + S2-[HS

= 9.1×10-5 M

[H+] = 9.54×10-5 M

Ka =$$\dfrac{ ([H^{+}][S^{2-}])}{[HS^{–}] }$$

= 1.2×10-13 (given)

Ka = $$\dfrac{ (9.1× 10^{-5} ×( S^{2-}))}{9.1×10^{-5}}$$

[S2-] = 1.2×10-13 M

Case 2 – HCl is present

[HS] = 9.1×10-8 M

[H+] = 0.1 M

Ka = 1.2×10-13 M

=$$\dfrac{0.1×S^{2-}}{9.1×10^{-8}}$$

Therefore, [S2-

= 1.092×10-19 M

Answered by Abhisek | 1 year ago

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