Suppose 0.10 mol of cu(no_3)_2 and 1.50 mol of nh_3 are dissolved in water and diluted to a total volume of 1.00 l. calculate the concentrations of cu(nh_3)_4^2+ and of cu^2+ at equilibrium.

Explanation:

It is known that the coefficients change in concentration and in the exponents.  Hence, the reaction equation will be as follows.

According to the ICE table,

Initial :    0.10         1.50                        0

Change : -x         -4x                        +x

Equilibrium: 0.10 - x  1.50 - 4x            x  

Hence, the mass action expression is as follows.

                        =

As, the value of is huge, it means that the reaction is very product favored. Hence, we need to find the limiting reactant first and then  we get to know what x should be.

In the given reaction ammonia is the limiting reactant, because there is less than 4 times the ammonia as the copper cation. Thus, we expect it to run out first, and so, x is approximately equal to 0.25 M.

So, putting the given values into the above equation as follows.

                              =

From here

                       = M

Therefore, we can "re-solve" for x to get and verify that it is still ≈0.250 M.

                x =

                  =

                    = 0.37491425 M

Thus, we can conclude that concentration of () is  0.37491425 M.