Question

If 2.35 g of CuSO₄ is dissolved in 9.07 × 10² mL of 0.360 M NH₃, calculate the concentrations of the following species at equilibrium.

Cu²⁺

_____×10 __M Enter your answer in scientific notation.

NH₃

_{________} M

Cu(NH₃)₄²⁺

^_________M

Answer 1

CuSO4 + 4 NH3 ------> Cu(NH3)4 + SO4

Find out how many moles of CuSO4 you have. Assuming the CuSO4 is anhydrous.

CuSO4 molecular weight is 159.62 g/mol.

=(2.35 g)/(159.62 g/mol)

= 0.01472 mol CuSO4

Next, figure out how many moles of NH3 you have:

0.360 M = (? moles NH3)/(0.907 L)

= 0.326556 mol NH3

You need 4 equivalents of NH3 for every equivalent of Cu2+ ion (the SO4 is just a spectator ion, so we'll ignore it).

1 equiv. = 0.01472 moles of Cu2+
4 equiv. = 0.05888 moles of NH3 needed

So the Cu2+ is the limiting reagent.

This means you will use up all of the Cu2+ ion, generating 0.01472 moles of Cu(NH3)4.

You will use up 0.05888 moles of NH3, so we will subtract that from the 0.326556 moles of NH3 that we started with:

0.326556 - 0.05888 = 0.267676 moles of NH3 left

Now we can figure out final concentrations:

[Cu2+] = 0 moles/0.9071 L = 0 M

[NH3] = 0.267676 moles/0.9071 L = 0.2950 M

[Cu(NH3)4] = 0.01472 moles/0.9071L = 0.0162 M

This of course assumes all of the Cu2+ ion is complexed by the NH3, which is should be given the large excess of NH3 around (given there is nearly 5 times as much NH3 around as is needed to complex the Cu2+ ions).