Question

Balance equation for the following reaction:
 Copper sulphate solution is added to sodium hydroxide solution.
A.\[{\text{CuS}}{{\text{O}}_4}{\text{ }} + {\text{ NaOH}}\; \to {\text{ Cu}}{\left( {{\text{OH}}} \right)_2}{\text{ }} + {\text{ N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}\]
B.\[{\text{2CuS}}{{\text{O}}_4}{\text{ }} + {\text{ NaOH}}\; \to {\text{ Cu}}\left( {{\text{OH}}} \right){\text{ }} + {\text{ N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}\]
C.\[{\text{CuS}}{{\text{O}}_4}{\text{ }} + {\text{ }}2{\text{NaOH}}\; \to {\text{ Cu}}{\left( {{\text{OH}}} \right)_2}{\text{ }} + {\text{ N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}\]
D.\[2{\text{CuS}}{{\text{O}}_4}{\text{ }} + {\text{ }}2{\text{NaOH}}\; \to {\text{ }}2{\text{Cu}}{\left( {{\text{OH}}} \right)_2}{\text{ }} + {\text{ N}}{{\text{a}}_3}{\text{S}}{{\text{O}}_4}\]

Answer 1

Hint:During a chemical reaction net charge (either positive or negative) remains the same before & after the reaction. Check the number of atoms and oxidation states on both sides.

Complete step by step solution:
During a chemical reaction, sum of charge on all cation or anion on reactant is equal to sum of charge of all cations or anion on product side. That is, net positive or negative charge remains the same before and after the reaction in a balanced chemical equation.
As copper sulphate solution is added to sodium hydroxide solution, forming products that are copper hydroxide and sodium sulphate. As copper solutions have cupric cation and sulphate anion. On the other hand sodium hydroxide solution has sodium cation and hydroxide anion. During mixing, cation of one solution mixes with anion of another solution and vice versa. So in our case, cupric ion from copper sulphate solution reacts with hydroxide ion of sodium hydroxide solution and results into formation of copper hydroxide in order to make compound, two hydroxide ion net charge - 2 combine with one cupric ion to keep neutrality of compound. Similarly sodium ion from sodium hydroxide will react with sulphate ion of copper sulphate and results in formation of sodium sulphate as two sodium ion will combine with one sulphate ion in order to keep neutrality of species.
Thus overall reaction will be: \[{\text{CuS}}{{\text{O}}_4}{\text{ }} + {\text{ NaOH}}\; \to {\text{ Cu}}{\left( {{\text{OH}}} \right)_2}{\text{ }} + {\text{ N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}\]
After balancing net reaction will be: \[{\text{CuS}}{{\text{O}}_4}{\text{ }} + {\text{ }}2{\text{NaOH}}\; \to {\text{ Cu}}{\left( {{\text{OH}}} \right)_2}{\text{ }} + {\text{ N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}\]
Net positive charge on reactant plus 4 which is equal to net positive charge on products that is plus 4 and same for negative charge.

Thus, the correct option is C.

Note:
 The reaction which involves exchange of ions does not involve change in oxidation state of ions but oxidation state of ion changes in redox reactions. Reduction causes decrease in oxidation number and oxidation causes increase in oxidation number. But still net charge remains the same even in redox reactions.