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An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation of HS from
H2S is 1.0×10−7 and that of S2− from HS ions is 1.2×10−13 then the concentration of S2− ions in aqueous solution is :

  • Option 1)

    5\times 10^{-19}

  • Option 2)

    5\times 10^{-8}

  • Option 3)

    3\times 10^{-20}

  • Option 4)

    6\times 10^{-21}

 

Answers (1)

As we lerned

Ionization reaction of dibasic acid -

H_{2}X(aq)\rightleftharpoons H^{+}(aq)+H\bar{X}(aq)


\bar{H}X(aq)\rightleftharpoons H^{+}(aq)+X^{2-}(aq)

- wherein

K_{a1}=\frac{[H^{+}]\:[\bar{H}X]}{[H_{2}X]}


K_{a2}=\frac{[H^{+}]\:[X^{2-}]}{[{H}X^{-}]}


K_{a1}\:and\:K_{a2}   are called first and second ionization constant.

 

 

t = 0        0.1          0.2

t = eq    0.1-x         0.2+2x    x    (x<<0.1)

\therefore \frac{\left [ H^{+} \right ]^{2}\left [ S^{-2} \right ]}{\left [ H_{2}S \right ]}=K_{eq}\; \; \therefore \frac{(0.2)^{2}x}{0.1}=1.2\times 10^{-20}

\therefore x=\frac{1.2\times10^{-20}\times 0.1 }{0.04}=3\times 10^{-20}\\*

\therefore \left [ S^{-2} \right ]=3\times 10^{-20}

 


Option 1)

5\times 10^{-19}

Option 2)

5\times 10^{-8}

Option 3)

3\times 10^{-20}

Option 4)

6\times 10^{-21}

Posted by

Vakul

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