Question

The concentration of $N{a}^{+}$ in a solution prepared by mixing $30.0$ mL of $0.12M$ $NaCl$ with $70$ mL of $0.15$ M $N{a}_{2}S{O}_4$ is :

• A. $0.135M$
• B. $0.141M$
• C. $0.210M$
• D. $0.246M$

Answer 1

The correct option is D $0.246M$

The dissociation reactions are as follows:
$NaCl\leftrightharpoons N{a}^{+}+C{l}^{-}$
$N{a}_{2}S{O}_4\leftrightharpoons 2N{a}^{+}+{S{O}_4}^{2-}$
The expression for the molarity of the mixture is $[N{a}^{+}{]}_{mix}=\frac{M_1{V}_1+M_2{V}_2}{V_1+V_2}$
Substituting values in the above expression, we get
$[N{a}^{+}{]}_{mix}=\frac{0.12\times 30+0.30\times 70}{30+70}=0.246M$