# Question 4045b

May 16, 2015

The equilibrium concentration of ${S}^{2 -}$ is $5.5 \cdot {10}^{- 7} \text{M}$.

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

$\text{ } H {S}_{\left(a q\right)}^{-} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + {S}_{\left(a q\right)}^{2 -}$
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 = ([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

${K}_{c} = {x}^{2} / \left(1.00\right) = 3 \cdot {10}^{- 13} \implies x = 5.5 \cdot {10}^{- 7}$

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

$\left[{S}^{2 -}\right] = \textcolor{g r e e n}{5.5 \cdot {10}^{- 7} \text{M}}$