Question #4045b

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Question #4045b

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Answer 1 · 2015-05-16T09:35:31

The equilibrium concentration of $S^{2 -}$ $S^(2-)$ is $5.5 \cdot 10^{- 7} M$ $5.5 * 10^(-7)"M"$ .

All you really have to do to solve this problem is use an ICE table for the equilibrium reaction given. The initial concentrations of $H^{+}$ $H^(+)$ and $S^{2 -}$ $S^(2-)$ will be zero.

$H S_{(a q)}^{-} ⇌ H_{(a q)}^{+} + S_{(a q)}^{2 -}$ $" "HS_((aq))^(-) rightleftharpoons H_((aq))^(+) + S_((aq))^(2-)$
I....1.00................0.............0
C....(-x)................(+x).........(+x)
E...1.00-x.............x..............x

By definition, the equilibrium constant for this reaction will be

$K_{c} = \frac{[H^{+}] \cdot [S^{2 -}]}{[H S^{-}]} = \frac{x \cdot x}{1.00 - x} = \frac{x^{2}}{1.00 - x}$ $K_c = ([H^(+)] * [S^(2-)])/([HS""^(-)]) = (x * x)/(1.00 - x) = x^2/(1.00 - x)$

Since the equilibrium constant is so small, you can approximate (1.00 - x) with 1.00. This will get you

$K_{c} = \frac{x^{2}}{1.00} = 3 \cdot 10^{- 13} \Rightarrow x = 5.5 \cdot 10^{- 7}$ $K_c = x^2/(1.00) = 3 * 10^(-13) => x = 5.5 * 10^(-7)$

Since $x$ $x$ represents the equilibrium concentration of both $H^{+}$ $H^(+)$ and $S^{2 -}$ $S^(2-)$ , the answer will be

$[S^{2 -}] = 5.5 \cdot 10^{- 7} M$ $[S^(2-)] = color(green)(5.5 * 10^(-7)"M")$

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Question #4045b

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