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I read that:

Take a cupful of water in a beaker and add a few drops of dilute sulfuric acid. Heat the water. When it starts boiling add copper sulfate powder slowly while stirring continuously. Continue adding copper sulfate powder till no more powder can be dissolved. Filter the solution. Allow it to cool. Do not disturb the solution when it is cooling. Look at the solution after some time. Can you see the crystals of copper sulfate? If not, wait for some more time.

So why is sulfuric acid required in this experiment?

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The sulfuric acid will dissolve any residual copper metal or any copper oxides that exist as impurities in the copper sulfate, converting those impurities into copper sulfate.

$$\ce{H2SO4(aq) + Cu(s) -> CuSO4(aq) + H2(g)}$$ $$\ce{H2SO4(aq) + CuO(s) -> CuSO4(aq) + H2O(l)}$$ $$\ce{2H2SO4(aq) + Cu2O(s) -> 2CuSO4(aq) + H2O(l) +H2(g)}$$

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Even if no impurities are present in your copper sulfate, your water might be slightly basic. (See this question.) Also water typically has dissolved $\ce{CO2}$, which in neutral solution means some $\ce{HCO3-}$.

By lowering the pH throught he addition of dilute sulfuric acid you ensure that the concentration of $\ce{OH-}$ and $\ce{HCO3-}$ is extremely low. Either of these ions, if present in significant amounts, will cause some coprecipitation (since both copper hydroxide and copper carbonates are much less soluble than copper sulfate) and as a result you wouldn't get nice big crystals of pure $\ce{CuSO4\cdot 5H2O}$.

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    $\begingroup$ I do think the acid is to get rid of the CO2 species. It isn't a problem of coprecipitation but rather nucleation. Any species of copper carbonate precipitating would form nucleation sites for the copper sulfate. So instead of getting a few large crystals forming slowly you'd get a zillion small crystals. $\endgroup$
    – MaxW
    Sep 24, 2016 at 16:19

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