Question

$...{Na}_3{PO}_4\left(aq\right)+...{Mg\left({NO}_3\right)}_2\left(aq\right)\to ...{{Mg}_3\left({PO}_4\right)}_2\left(s\right)+...Na{NO}_3\left(aq\right)$
Equal volume of 0.10 M sodium phosphate and 0.10 M magnesium nitrate are mixed in a container according to the above unbalanced reaction equation. After the completion of the reaction, what is the concentration of sodium ions in the solution?

• A. $\sim 0M$, because all the sodium ions will precipitate
• B. 0.05 M
• C. 0.10 M
• D. 0.15 M
• E. 0.30 M

Answer 1

The correct option is D 0.15 M

After the completion of the reaction, the concentration of sodium ions in the solution is $\left[N{a}^{+}\right]=\frac{3\times 0.10}{2}=0.15$ M.
In the above expression, the concentration of sodium phosphate (0.10 M) is multiplied with 3 so that each molecule of sodium phosphate on dissociation gives 3 sodium ions.
Also the value is divided with 2 because when equal volumes of two solutions are mixed, the concentration of each species is doubled.