Question

If 2.50 g of CuSO₄ is dissolved in 8.13 x 10² ml of 0.310 M NH_3, calculate the concentrations of the following species at equilibrium.

Cu(NH₃)₄²⁺? __ M

NH₃? __ M

Cu²⁺? _____ x 10^??

enter in scientific notation

Answer 1

Solution :-

Lets first calculate the moles of the CuSO4

Moles of CuSO4 = mass / molar mass

                              = 2.50g /159.6 g per mol

                             = 0.01566 mol CuSO4

So the concentration of the Cu^2+ = 0.01566 mol / 0.813 L = 0.01926 M

Now lets calculate the moles of the NH3

Moles of NH3 = molarity * volume in liter

                          = 0.310 mol per L * 0.813 L

                         = 0.252 mol NH3

1 mol Cu^2+ need 4 mol NH3 therefore

0.01566 mol Cu^2+ * 4 mol NH3 / 1 mol Cu^2+ = 0.06264 mol NH3

So the moles of NH3 remain = 0.252 mol – 0.06264 mol = 0.18936 mol NH3

So the concentration of the NH3 remain = 0.18936 mol / 0.813 L = 0.2329 M

Kf of the Cu(NH3)4^2+ = 5.6*10^11

Since the formation constant is very high so all of the Cu^2+ is converted into the complex

So the concentration of the complex formed = 0.01566 mol Cu(NH3)4^2+ mol / 0.813 L = 0.019262 M

Now lets use this concentration and set up the ice table

Cu(NH3)4^2+         ---------- > Cu^2+               4NH3

0.019262                                     0                         0.2329

-x                                                  +x                       +4x

0.019262 –x                                x                        0.2329 +4x

Now lets write the equilibrium constant equation

1/Kf = [Cu^2+] [NH3]^4 /[Cu(NH3)4^2+]

1/5.6*10^11 = [x][0.2329+4x]^4 /[0.019262-x]

1.7857*10^-12 = [x][0.2329+4x]^4 /[0.019262-x]

1.7857*10^-12 * [0.019262-x] = [x][0.2329+4x]^4

Solving for the x we get

1.17*10^-11 = x

Therefore the concentration of the species at the equilibrium is as follows

[Cu(NH3)4^2+] = 0.01962 M

[NH3] = 0.2329 M

[Cu^2+] = 1.17*10^-11 M