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Q.

Equilibrium const.
$H_{2} S (a q) ⇌ H^{+} (a q) + {HS}^{-} (a q) K_{1} = 9.5 \times 10^{- 8}$
${HS}^{-} (a q) ⇌ H^{+} (a q) + S^{2 -} (a q) K_{2} = 1.0 \times 10^{- 19}$
What is the equilibrium constant for the reaction?
$S^{2 -} (aq) + 2 H^{+} (aq) ⇌ H_{2} S (aq) K = ?$

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a

$9.7 \times 10^{- 14}$

b

$1.0 \times 10^{26}$

c

$9.5 \times 10^{11}$

d

$9.5 \times 10^{- 27}$

answer is D.

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Detailed Solution

The equilibrium constant can be calculated as;

$K = \frac{1}{K_{1} \times K_{2}}$

$K = \frac{1}{9.5 \times 10^{- 8} \times 1.0 \times 10^{- 19}}$

$K = 1 \times 10^{26}$

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Equilibrium const.H2S(aq)⇌H+(aq)+HS−(aq) K1=9.5×10−8HS−(aq)⇌H+(aq)+S2−(aq) K2=1.0×10−19What is the equilibrium constant for the reaction?S2−(aq)+2H+(aq)⇌H2S(aq) K=?