Question

Why is dilute sulphuric acid added during crystallization of copper sulphate?
A) Prevent hydrolysis of copper sulphate
B) Prevent crystallization of copper sulphate
C) Enhance formation of large crystals
D) All of the above

Answer 1

Sulphuric acid is mainly used to lower the pH of the solution and make it acidic. While preparation of copper sulphate solution a little amount of dil. Sulphuric acid is added. Copper sulphate powder is dissolved until no more powder can be dissolved. This is called the supersaturated solution of copper sulphate.

Complete Step By Step Answer:
The above solution is prepared as follows: In a cupful of water in a beaker, add a few drops of sulphuric acid and heat it. When the water is near to boiling, add copper sulphate powder and slowly stir. Keep dissolving the solute until no more solute can be dissolved. Filter the solution and cool it down. After looking you can be copper sulphate crystals being separated out.
The sulphuric acid will dissolve any residual copper metal or its oxides present in the solution as impurities and convert those to copper sulphate as well. The reaction that takes place is:
${H_2}S{O_{4(aq)}} + C{u_{(s)}} \to CuS{O_{4(aq)}} + {H_{2(g)}}$
${H_2}S{O_{4(aq)}} + Cu{O_{(s)}} \to CuS{O_4} + {H_2}{O_{(l)}}$
$2{H_2}S{O_{4(aq)}} + C{u_2}{O_{(s)}} \to 2CuS{O_{(aq)}} + {H_2}{O_{(l)}} + {H_{2(g)}}$
In water, when copper sulphate is dissolved, it tends to form a complex with water by hydrolysis. Hence to prevent that a small amount of sulphuric acid is added.
Hence the correct answer is Option (A).

Note:
Even if there are no impurities present, the water is always slightly basic. It has dissolved carbon dioxide which forms $HCO_3^ -$ ions. By adding sulphuric acid the pH is lowered such that the concentration of $O{H^ - }\& HCO_3^ -$ is extremely low. The presence of these ions causes coprecipitation of copper hydroxide and copper carbonates which are much less soluble in water. Therefore, by addition of sulphuric acid we get pre crystals of $CuS{O_4}.5{H_2}O$ .