Question

Suppose 40.8g of copper(II) acetate is dissolved in 200.mL of a 0.70 M aqueous solution of sodium chromate.

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