Question #4045b

1 Answer
Stefan V.
May 16, 2015

The equilibrium concentration of #S^(2-)# is #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^(+)# and #S^(2-)# will be zero.

#" "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 = ([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/(1.00) = 3 * 10^(-13) => x = 5.5 * 10^(-7)#

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

#[S^(2-)] = color(green)(5.5 * 10^(-7)"M")#

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