The first ionization constant of H_{2}S 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 H_{2}S is 1.2 × 10^{–13}, calculate the concentration of S^{2–} under both conditions.

Asked by Pragya Singh | 1 year ago | 111

To calculate [HS^{–}]

To find [HS^{–}]

Case 1 – HCl is absent.

Now

K_{a} = \( \dfrac{ ([H^{+}][HS^{–}])}{[H_2S] }\)

= 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}

x^{2} = 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,

K_{a} = \( \dfrac{(H+HS^{–})}{H_2S}\)

= (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 [S^{2-}]:

Case 1 – HCl is absent.

The dissociation of HS^{–} is given by the equation:

HS^{–} ⇌ H^{+} + S^{2-}[HS^{–}]

= 9.1×10^{-5} M

[H^{+}] = 9.54×10^{-5} M

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

= 1.2×10^{-13} (given)

K_{a} = \(
\dfrac{ (9.1× 10^{-5} ×( S^{2-}))}{9.1×10^{-5}}\)

[S^{2-}] = 1.2×10^{-13} M

Case 2 – HCl is present

[HS^{–}] = 9.1×10^{-8} M

[H^{+}] = 0.1 M

K_{a} = 1.2×10^{-13} M

=\( \dfrac{0.1×S^{2-}}{9.1×10^{-8}}\)

Therefore, [S^{2-}]

= 1.092×10^{-19} M

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