# How can you find the concentration of copper(II) ions in a solution of 0.5 mole copper(II) chloride and 0.5 mole of copper(II) sulfate?

The question I'm trying to answer is below and the answer is C. I tried finding the total moles of copper(II) ions first by using mole ratios for copper(II) sulfate and copper(II) chloride, obtaining an answer of 0.42 moles. Then, I used $n=cv$ and obtained the concentration, using the volume in the question in litres, of 0.84, which is incorrect. Now, I'm sure my method incorrect but I can't think of another way to this.

• Where are you getting your numbers from? They told you that 0.5 mol of $\ce{CuCl_2}$ and 0.5 mol of $\ce{CuSO_4}$ were placed together. That means you have 0.5 mol + 0.5 mol = 1 mol of copper ions. Divide that by the volume, which is given to you as 0.500 L, giving you 2.0 mol/L. Your numbers don’t seem to match up with the question at all. – lightweaver Apr 3 '16 at 13:54
• @lightweaver Yes but the number of moles in the question are the moles of the two solutions, not the copper ions, I believe. – user307397 Apr 3 '16 at 14:09
• @lightweaver I gor my number by using the ratio 1mol copper(ii):2mol chlorine and 1mol copper(ii):1 mol $So_4$ – user307397 Apr 3 '16 at 14:13
• The question is: "What is the concentration of $\ce{Cu^{2+}_{(aq)}}$ ions in the solution in $mol\,l^{-1}$. – aventurin Apr 3 '16 at 14:21
• 1 mole copper(II) sulfate contains 1 mole $Cu^{2+}$, and 1 mole copper(II) chloride also contains 1 mole $Cu^{2+}$. – aventurin Apr 3 '16 at 15:40

1 mole of copper(II) sulphate contains 1 mole of copper ($\ce{Cu^{2+}}$) and one mole of sulphate ($\ce{SO_4^{2-}}$). Or, if you want, 1 mole of copper, 1 mole sulphur, and 4 moles of oxygen.
1 mole of copper ($\ce{Cu^{2+}}$) in $\mathrm{0.5\,l}$ water gives us a concentration of $\mathrm{2\, mol\,l^{-1}}$