Question

In the laboratory a student combines 40.6 mL of a 0.113 M copper(II) sulfate solution with 26.4 mL of a 0.329 M copper(II) iodide solution. What is the final concentration of copper(II) cation?

Answer 1

Answer : The final concentration of copper(II) ion is, 0.198 M

Explanation :

First we have to calculate the moles of $CuSO_4$ and $CuI_2$.

$\text{Moles of }CuSO_4=\text{Concentration of }CuSO_4\times \text{Volume of solution}$

$\text{Moles of }CuSO_4=0.113mol/L\times 0.0406L=0.00459mol$

Moles of $CuSO_4$ = Moles of $Cu^{2+}$ = 0.00459 mol

and,

$\text{Moles of }CuI_2=\text{Concentration of }CuI_2\times \text{Volume of solution}$

$\text{Moles of }CuI_2=0.329mol/L\times 0.0264L=0.00869mol$

Moles of $CuI_2$ = Moles of $Cu^{2+}$ = 0.00869 mol

Now we have to calculate the total moles of copper(II) ion and total volume of solution.

Total moles copper(II) ion = 0.00459 mol + 0.00869 mol

Total moles copper(II) ion = 0.0133 mol

and,

Total volume of solution = 40.6 mL + 26.4 mL = 64.0 mL = 0.067 L    (1 L = 1000 mL)

Now we have to calculate the final concentration of copper(II) ion.

$\text{Final concentration of copper(II) ion}=\frac{\text{Total moles}}{\text{Total volume}}$

$\text{Final concentration of copper(II) ion}=\frac{0.0133mol}{0.067L}=0.198mol/L=0.198M$

Thus, the final concentration of copper(II) ion is, 0.198 M