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* 10^(-4)g`

Volume of solution = 2.25 L

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

`Concentration=(Mass)/(Volume)=(2.85* 10^(-4)g)/(2.25L)=1.27* 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