Question

The concentration of $S{O}_4^{2-}$ ion in $III$ compartment will be:

• A. $1.0M$
• B. $1.025M$
• C. $0.95M$
• D. $0.975M$

Answer 1

Answer: (D)

$0.975M$

Equivalent of $Z{n}^{2+}$ produced = $0.1$ or moles of $Z{n}^{2+}=\frac{0.1}{2}\Rightarrow 0.05$
+ve charge increases in first compartment so due to interaction and maintain electrical neutrality $Z{n}^{2+}$ move toward II compartment and $N{O}_3^{-}$ move towards first compartment. Solution is always electrically neutral so charge of $1Z{n}^{2+}$ is neutralized by $2N{O}_3^{-}$.
$\therefore \left[Z{n}^{2+}\right]$ in first compartment
$=1+\frac{0.05}{2}=1.025M$
Concentration of $N{O}_3^{-}$ in second compartment
$=1-0.05=0.95M$
In third compartment moles of $C{u}^{2+}$ reduced
$=\frac{0.05}{2}=0.025$
Relatively -ve charge increased so $S{O}_4^{2-}$ and $N{a}^{+}$ move toward opposite direction to maintain electrical neutrality
$[S{O}_4^{2-}{]}_{remaining}=1-\frac{0.025}{2}\Rightarrow 0.975M$ {Option D}