# CuCl2 dissolved in water

We dissolve $10\ \mathrm{g}$ of $\ce{CuCl2}$ in $0.1\ \mathrm{L}$ of water.
Given that:

$M(\ce{Cl})=35.5\ \mathrm{g/mol}$
$M(\ce{Cu})=63.5\ \mathrm{g/mol}$

How can one calculate the molar concentration of $\ce{Cu^2+}$ and $\ce{Cl-}$?

First of all I calculated the the quantity of matter of $\ce{CuCl2}$:

$n(\ce{CuCl2})=0.074\ \mathrm{mol}$

The the molar concentration:

$c=\frac{0.074\ \mathrm{mol}}{0.1\ \mathrm{L}}=0.74\ \mathrm{mol/L}$

Then I don’t know how can I calculate the molar concentration of $\ce{Cu^2+}$ and $\ce{Cl-}$ because I don’t know the quantity of matter of them!

But here im blocked!

• You can assume that CuCl2 will be fully dissociated. Jun 1 '15 at 9:00
• @JaroslavKotowski yes can this change something ? Jun 1 '15 at 9:03

Jaroslav already gave you a good hint: Assume full dissociation, which leads to the following equation: $$\ce{CuCl2 <=> Cu^2+ + 2Cl^-}$$

Since you know the initial concentration of copper chloride, $c(\ce{CuCl2})=0.74~\mathrm{mol/L}$, what can you say about the ratio of $n(\ce{CuCl2})/n(\ce{Cu^2+})$ and $n(\ce{CuCl2})/n(\ce{Cl^-})$?

Hint: In one mole of of a salt of the composition $\ce{AB}$, you will find one mole of $\ce{A+}$ and one mole of $\ce{B-}$, according to $\ce{AB <=> A+ + B-}$.

Therefore:

$n(\ce{CuCl2})/n(\ce{Cu^2+}) = 1$, thus $c(\ce{CuCl2})=n(\ce{Cu^2+})=0.74~\mathrm{mol/L}$
and
$n(\ce{CuCl2})/n(\ce{Cl^-}) =\frac12$, thus $c(\ce{CuCl2})=2\cdot c(\ce{Cl^-})=1.48~\mathrm{mol/L}$

• I think that $n(\ce{CuCl2})/n(\ce{Cu^2+})$ is the concentration of Cu^2+ ?! Jun 1 '15 at 9:26
• @user233658 That ratio is just a number, check the part in the spoiler tags, I included the solution. Jun 1 '15 at 9:29
• How did you know that his ratio is 1 ? Is it because there is 2 molecule of Cu^2+ ? n(Cu^2+)=0.074mol ? How did you know it :/ ? Oh im feeling stupid :( Jun 1 '15 at 9:33
• The formula gives you the ratio already. If you assume a molecule $\ce{CuCl2}$ there is one copper and two chlorine. So for each mole of copper chloride, full dissociation will give you one mole of copper ions. Jun 1 '15 at 9:36

When a salt is fully dissociated, than the atoms in a solution exist separately: $$\ce{CuCl2<=>Cu^{2+}}+2\ce{Cl-}$$ Amount of substance is 0.074 mol. The molecule is made of 1 atom of Cu and 2 atoms of Cl.

$$n=\frac{m}{M}$$ $$n \cdot M =m$$ Amount of copper in the mixture then is: $$0.074~\mathrm{mol} \cdot 63.5~\mathrm{g/mol} = 4.7~\mathrm{g}$$ and the rest is formed by chlorine. $$10~\mathrm{g}- 4.7~\mathrm{g} = 5.3~\mathrm{g}$$