The formation constant of ${\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{+2}$ is $6 \times {10}^8$ at ${25}^0\mathrm{C}.$  If 50 ml of 2.0 M  is ${\mathrm{NH}}_3$added to 50 ml of 0.20 M solution of ${\mathrm{Ni}}^{2+}$ ion will be nearly equal to

50ml of 2M NH₃ + 50ml of 0.2M Ni⁺² solution

Hence ${\mathrm{n}}_{{\mathrm{Ni}}^{+2}}=\frac{50\times 0.2}{100}=0.1\mathrm{mole} ; {\mathrm{n}}_{{\mathrm{NH}}_3}=\frac{50\times 2}{100}=1\mathrm{mole}$

 ${\mathrm{Ni}}^{2+} + 6{\mathrm{NH}}_3 \to {\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{+2}; {\mathrm{K}}_{\mathrm{f}} = 6 \times {10}^8$

 0.1 mole        1 mole                  0

   0.1                  1-0.6                 0.1

${\mathrm{K}}_{\mathrm{f}}= \frac{\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6^{+6}\right]}{\left[{\mathrm{Ni}}^{+2}\right] {\left[{\mathrm{NH}}_3\right]}^6}$

$= \frac{\left(0.1\right)}{\left[{\mathrm{Ni}}^{+2}\right] {\left(0.4\right)}^6} = 6 \times {10}^8$

$\left[{\mathrm{Ni}}^{2+}\right] =$$4 \times {10}^{-8}$