- #1

brake4country

- 216

- 7

## Homework Statement

Which of the following expressions represents the solubility product for Cu(OH)

_{2}?

(A) K

_{sp}=[Cu

^{2+][OH-}]

^{2}

(B) K

_{sp}=[Cu

^{2+}]2[OH

^{-}]

(C) K

_{sp}=[Cu

^{2+}]2[OH

^{-}]

^{2}

(D) K

_{sp}=[Cu

^{2+}][OH

^{-}]

## Homework Equations

K

_{sp}= [A][ B]

## The Attempt at a Solution

Okay, so I understand equilibrium expressions and in this case, solids and pure liquids are left out of the expression. My question is regarding the coefficients and exponents of solubility expressions. In chemical equilibrium expression is the concentration of the products are over the reactants (with exponents being the coefficients of each substance). Knowing that solubility rules are pretty much the same, I know that answer A is correct but my books reasoning for the following reaction does not make sense:

BaF

_{2}↔Ba

^{2+}

_{aq}+2F

^{-}

_{aq}

=K

_{sp}=[Ba

^{2+}][F

^{-}]

^{2}

However, they write the equation as:

2.4 x 10

^{-5}= (x)(2x)

^{2}(and then they solve for x).

This expression is not consistent, say, with answer choice A in another problem. How did they get 2x in this instance for BaF

_{2}? Shouldn't the first problem above be K

_{sp}=[Cu

^{2+}][2OH]

^{2}?[/B][/B][/B]

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