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Question

Copper(II) ions can be precipitated as copper(II) carbonate.

CuSO4(aq) + Na2CO3(aq) → CuCO3(s) + Na2SO4(aq)

How many mL of 0.670 M Na2CO3 would be needed to precipitate all of the copper ions in 74.5 mL of 0.419 M CuSO4?

125

Incorrect mL

Answer 1

Since most molecules that involve a metal, such as copper, bound to a carbonate group are highly insoluble in water, we will be working under the assumption in this scenario that 100% of the theoretical copper (II) carbonate yield will form. Thus, in this case, our first step is to check to make sure that our stoichiometric coefficients are balanced. Luckily, they are here, so our balanced chemical equation remains:

CuSO₄(aq) + Na₂CO₃(aq) → CuCO₃(s) + Na₂SO₄(aq)

From here, we can then calculate the amount of moles of copper ions in 74.5 mL of .419 M CuSO₄. To do this, we multiply out the volume in Liters by the molarity like so:

moles of CuSO₄ = (.0745 Liters of CuSO₄)*(.419 M)

moles of CuSO₄ = .0312 moles

Since our balanced reaction shows that for every 1 mole of CuSO₄ used, 1 mole of Na₂CO₃ is also used, we do not need to multiply our answer by anything to account for a change in molar equivalents of reactants and need .0312 moles of Na₂CO₃ as well. Using the same equation we used before, we can then calculate the volume of NaCO₃ needed to complete this reaction fully:

moles of Na₂CO₃ = (Volume of NaCO₃)*(Molarity of Na₂CO₃)

.0312 = (Volume of Na₂CO₃)*(.670)

Volume of Na₂CO₃ = .0466 L

Volume of Na₂CO₃ = 46.6 mL