Question

What volume of 0.416 M Mg(NO3)2 should be added to 255 mL of 0.102 M KNO3 to produce a solution with a concentration of 0.278 M NO3- ions ? (asume that volumes are additive)

Answer 1

Given:

Molarity of  $Mg(NO_3)_2$ = 0.146 M

Molarity of $KNO_3$ = 0.102 M

Volume of $KNO_3$ = 225 ml

Final concentration of $NO_3^-$ ions = 0.278 M

To Find :

Volume of $Mg(NO_3)_2$ required = ?

Solution :

Since we know that molarity is given as :

$M = \frac{no \hspace3 of \hspace3 moles \hspace3 of \hspace3 solute }{volume \hspace3 of \hspace3 solution (l)}$

As, $Mg(NO_3)^2$ dissociates completely , so we have :

$Mg(NO_3)^2 \rightarrow Mg^{2+} + 2NO_3^-$

And also ;

$KNO_3 \rightarrow K^+ + NO_3^-$

Now final molarity of $NO_3^-$ ion =$\frac{n}{V}$

Or, M=$\frac{2M_1V_1 + M_2V_2}{V_1 + V_2}$

Where $M_1$ and $M_2$ are molarities of $Mg(NO_3)^2$ and $KNO_3$ respectively and similar for the volumes.

So,

$0.278 = \frac{2\times 0.416 \times V_1 + 0.102 \times 255 \times 10 ^{-3}}{V_1 + 255 \times 10^{-3}}$

Or, 0.278$V_1$ + 0.0708 = 0.832$V_1$ + 0.0212

Or, 0.0496 =$0.554 V_1$

Or, $V_1$ =0.089 $l$

Or, $V_1 = 89 \hspace2 ml$

So, the volume required is 89 ml.