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A solution contains 1.25E-6 M Cu2+(aq) and 1.25E-6 M OH- (aq). Will a precipitate form? The Ksp for Copper (II) Hydroxide is 4.80E-20 M3.
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Calculate Q and compare it to Ksp. First you need to know your reaction: \[Cu(OH)_2 \space (s) \leftarrow \rightarrow Cu^{2+} \space (aq) + 2 OH^- \space (aq)\] Now we can calculate Q: \[Q=[Cu^{2+}][OH^-]^2\]\[Q=(1.25 \times 10^{-6})(1.25 \times 10^{-6})^2\]\[Q=1.95 \times 10^{-18}\] In this case, Q > Ksp. That means the reverse reaction will be favored and a precipitate will form, in order to decrease the concentration of products and ultimately make Q=Ksp!
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