Question

Suppose 40.8g of copper(II) acetate is dissolved in 200.mL of a 0.70 M aqueous solution of sodium chromate.
Calculate the final molarity of copper(II) cation in the solution. You can assume the volume of the solution doesn't change when the copper(II) acetate is dissolved in it.
Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

0.42 M

Explanation:

The reaction that takes place is:

• Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)

First we calculate the moles of Na₂CrO₄, using the given volume and concentration:

(200 mL = 0.200L)

• 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄

Now we calculate the moles of Cu(CH₃COO)₂, using its molar mass:

• 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂

Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂:

• 0.224 mol - 0.14 mol = 0.085 mol

Finally we calculate the resulting molarity of Cu⁺², from the excess cations remaining:

• 0.085 mol / 0.200 L = 0.42 M