Ncert Exercises & Solutions

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.

H_{2} S \leftrightarrow H^{+} + {HS}^{-}

C_{i} 0.1 0 0

C_{f} 0.1 - x x x

Then, K_{a_{i}} = \frac{[H^{+}] [{HS}^{-}]}{[H_{s} S]}

9.1 \times 10^{- 8} = \frac{(x) (x)}{0.1 - x}

(9.1 \times 10^{- 8}) (0.1 - x) = x^{2}

Taking 0.1 - x M; 0.1 M, we have

(9.1 \times 10^{- 8}) (0.1) = x^{2}

9.1 \times 10^{- 9} = x^{2}

x = \sqrt{9.1 \times 10^{- 9}}

= 9.54 \times 10^{- 5} M

\Rightarrow [{HS}^{-}] = 9.54 \times 10^{- 5} M

Case ll (in the presence of HCl):

In the presence of 0.1 M of HCl, let

[H S^{-}]

be y M

Then, H_{2} S \leftrightarrow {HS}^{-} + H^{+}

C_{i} 0.1 0 0

C_{f} 0.1 - y y y

Now, Then, K_{a_{i}} = \frac{[{HS}^{-}] [H^{+}]}{[H_{s} S]}

9.1 \times 10^{- 8} = \frac{y \times 0.1}{0.1} (∵ 0.1 - y; 0.1 M)

(and 0.1 + y; 0.1 M)

9.1 \times 10^{- 8} = y

\Rightarrow [{HS}^{-}] = 9.1 \times 10^{- 8}

To calculate the concentration of

[S^{2 -}]

Case l (in the absence of 0.1 M HCl):

{HS}^{-} \leftrightarrow H^{+} + S^{2 -}

[{HS}^{-}] = 9.54 \times 10^{- 5} M (From first ionization, case I)

Let [S^{2 -}] be X .

Also, [H^{+}] = 9.54 \times 10^{- 5} M (From first ionization, case I)

\begin{array}{l} K_{a_{2}} = \frac{[H^{+}] [S^{2 -}]}{[{HS}^{-}]} \\ K_{a_{2}} = \frac{(9.54 \times 10^{- 5}) (X)}{9.54 \times 10^{- 5}} \\ 1.2 \times 10^{- 13} = X = [S^{2 -}] \end{array}

Case ll (in the presence of 0.1 M HCl):

Again, let the concentration of

H S^{-}

be X'M.

\begin{array}{l} [S^{2 -}] = X^{'} \\ Then, K_{a_{2}} = \frac{[H^{+}] [S^{2 -}]}{[{HS}^{-}]} \\ 1.2 \times 10^{- 13} = \frac{(0.1) (X^{'})}{9.1 \times 10^{- 8}} \\ 10.92 \times 10^{- 21} = 0.1 X^{'} \\ \frac{10.92 \times 10^{- 21}}{0.1} = X^{'} \\ X^{'} = \frac{1.092 \times 10^{- 20}}{0.1} \\ = 1.092 \times 10^{- 19} M \\ \Rightarrow K_{_{a_{i}}} = 1.74 \times 10^{- 5} \end{array}

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