Question

Determine the concentrations of MgCl2, Mg2+, and Cl− in a solution prepared by dissolving 2.852.85 × 10−4 g MgCl2 in 2.252.25 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm).

Answer 1

Answer : The concentration of $MgCl_2,Mg^{2+}\text{ and }Cl^-$ are 0.127 ppm, 0.127 ppm and 0.254 ppm respectively.

Explanation : Given,

Mass of $MgCl_2$ = $2.85\times 10^{-4}g$

Volume of solution = 2.25 L

First we have to calculate the concentration in terms of g/L.

$Concentration=\frac{Mass}{Volume}=\frac{2.85\times 10^{-4}g}{2.25L}=1.27\times 10^{-4}g/L=0.127mg/L$

conversion used : $1g=1000mg$

As we know that,

1 mg/L = 1 ppm

0.127 mg/L = 0.127 ppm

So, the concentration will be 0.127 ppm.

The concentration of $MgCl_2$ in ppm is, 0.127 ppm.

Now we have to calculate the concentration of $Mg^{2+}$ and $Cl^-$.

We assume that, $MgCl_2$ dissociates 100 % in the solution then the balanced reaction will be:

$MgCl_2(aq)\rightarrow Mg^{2+}(aq)+2Cl^-(aq)$

From the reaction we conclude that the mole ratio of $MgCl_2:Mg^{2+}:Cl^-$ is, 1 : 1 : 2. So,

The concentration of $Mg^{2+}$ = 0.127 ppm

The concentration of $Cl^-$ = 2 × 0.127 ppm = 0.254 ppm