# Why is sulfuric acid used in crystallization

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?

$$\ce{H2SO4(aq) + Cu(s) -> CuSO4(aq) + H2(g)}$$ $$\ce{H2SO4(aq) + CuO(s) -> CuSO4(aq) + H2O(l)}$$ $$\ce{2H2SO4(aq) + Cu2O(s) -> 2CuSO4(aq) + H2O(l) +H2(g)}$$
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 $\ce{CuSO4\cdot 5H2O}$.