Question

$Zn$ rod is placed in 100 L of 1M $\text{CuS}{\text{O}}_{\text{4}}$ solution so that molarity of $\text{C}{\text{u}}^{2+}$ changes to 0.7 M. The molarity of ${\text{SO}}_{\text{4}}^{\text{2 -}}$ at this stage will be:

• A. $0.8$ M
• B. 1 M
• C. $0.7$ M
• D. $1.8$ M

Answer 1

The correct option is B 1 M

The molarity of sulphate ion would be the same as that in the initial solution i.e. $1$M as it does not participate in the reaction. The reaction is as follow;

$Zn+C{u}^{2+}⟶Z{n}^{2+}+Cu$

The sulphate ion concentration would be the same before and after the reaction.

Hence, the correct option is $\left(B\right)$.