Question

A 130.0 mL sample of a solution that is 0.0126 M in NiCl2 is mixed with a 190.0 mL sample of a solution that is 0.400 M in NH3.

-After the solution reaches equilibrium, what concentration of Ni2+(aq) remains?

The value of Kf for Ni(NH3)62+ is 2.0×108.

Answer 1

Solution :-

Ni^2+    + 6NH3 ----- > Ni(NH3)6^2+

Moles of Ni^2+ = molarity * volume

                            =0.0126 mol per L * 0.130 L = 0.001638

Moles of NH3 = 0.400 mol per L * 0.190 L = 0.076 mol

Lets calculate the moles of NH3 needed to react with 0.001638 mol Ni^2+

0.001638 mol Ni^2+ * 6 mol NH3 / 1 mol Ni^2+ = 0.009828 mol

Therefore all the Ni^2+ is converted to the Ni(NH3)6^2+

Therefore the molarity og the Ni(NH3)6^2+ = 0.001638 mol / (0.130 L +0.190 L) = 0.00512 M

Ni^2+    + 6NH3 ----- > Ni(NH3)6^2+

0                     0                       0.00512 M

+x                  +6x                         -x

X                    6x                       0.00512-x

Kf=[ Ni(NH3)6^2+]/[Ni^2+][NH3]^6

2.0*10^8 = [0.00512-x]/[x][6x]^6

2.0*10^8 *46656x^7 = 0.00512-x

Solving for x we get

X=0.0047

Therefore the Ni^2+ remain after reaction = 0.0047 M