In an excess of NH3(aq), Cu2+ ion forms a deep blue complex ion, Cu(NH3)4^2+ (Kf=5.6x10^11). Calculate the concentration of Cu2+ in a solution adding 5.0x10^-3 mol of CuSO4 to .5 L of .4 M NH3
Cu^+2 + 4NH3 ==> Cu(NH3)4^+2
Kf = [Cu(NH3)4^+2]/([Cu^+2][NH3]^4
Substitute into the above (in M units) and solve for )Cu^+)
You can assume, with such a large constant, that [Cu(NH3)4^+2] = 0.01 and NH3 is about 0.4-(4*0.01) = about 0.36
So (Cu^+2) is about 1 x 10^-12M but you should confirm that.
To calculate the concentration of Cu2+ in the solution, we need to determine the amount of Cu2+ that reacts with NH3 to form the complex ion.
First, let's calculate the amount of Cu2+ from CuSO4 that reacts with NH3:
Amount of Cu2+ = 5.0x10^-3 mol CuSO4
Next, we need to determine the amount of NH3 that reacts with Cu2+:
Molarity of NH3 = 0.4 M NH3
Volume of NH3 = 0.5 L
The amount of NH3 can be calculated using the formula:
Amount of NH3 = Molarity x Volume
= 0.4 M NH3 x 0.5 L
= 0.2 mol NH3
Since the reaction between Cu2+ and NH3 is 1:4 (Cu2+: NH3), the amount of Cu2+ that reacts can be calculated by multiplying the amount of NH3 by the stoichiometric ratio:
Amount of Cu2+ used = 0.2 mol NH3 x (1 mol Cu2+/4 mol NH3)
= 0.2 mol NH3 x 1/4
= 0.05 mol Cu2+
Now, let's calculate the concentration of Cu2+ in the solution:
Volume of solution = 0.5 L
Concentration of Cu2+ = Amount of Cu2+ used / Volume of solution
= 0.05 mol Cu2+ / 0.5 L
= 0.1 M Cu2+
Therefore, the concentration of Cu2+ in the solution is 0.1 M.