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Dear Experts , I have a question How can CuO be synthesized from CuSO 4 and KOH? Plz reply asap.
Hi Disha, You'll form a blue, bulky precipitate of Cu(OH)2 first, according to CuSO4(aq) + 2 KOH(aq) -> K2SO4(aq) + Cu(OH)2(s) If you add too much KOH, the precipitate will dissolve (and you get an intensely blue solution of complex copper ions). Add it a little at a time until you see no more precipitate form. Filter off the copper(II) hydroxide, and wash it to remove the soluble potassium sulfate. Pour the clean precipitate into an evaporating dish and dry it very gently: Cu(OH)2(s) -> CuO(s) + H2O(g) You should see black powder form. That's the copper(II) oxide. If you heat it too much, it will decompose further into copper(I) oxide and oxygen.
Hi Disha,
You'll form a blue, bulky precipitate of Cu(OH)2 first, according to CuSO4(aq) + 2 KOH(aq) -> K2SO4(aq) + Cu(OH)2(s) If you add too much KOH, the precipitate will dissolve (and you get an intensely blue solution of complex copper ions). Add it a little at a time until you see no more precipitate form. Filter off the copper(II) hydroxide, and wash it to remove the soluble potassium sulfate. Pour the clean precipitate into an evaporating dish and dry it very gently: Cu(OH)2(s) -> CuO(s) + H2O(g) You should see black powder form. That's the copper(II) oxide. If you heat it too much, it will decompose further into copper(I) oxide and oxygen.
CuSO4(aq) + 2 KOH(aq) -> K2SO4(aq) + Cu(OH)2(s)
Filter off the copper(II) hydroxide, and wash it to remove the soluble potassium sulfate. Pour the clean precipitate into an evaporating dish and dry it very gently:
Cu(OH)2(s) -> CuO(s) + H2O(g)
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