The formation constant of $Ni {({NH}_{3})}_{6}^{2 +}$ is $6 \times 10^{8}$ at $25^{0} C$ . If $50 m l$ of $2.0 M$ ${NH}_{3}$ is added to $50 m l$ of $0.20 M$ solution of ${Ni}^{2 +}$ , the concentration of ${Ni}^{2 +}$ ion will be nearly equal to :

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$2 \times 10^{- 10}$ mole ${litre}^{- 1}$

$3 \times 10^{- 10}$ mole ${litre}^{- 1}$

$2 \times 10^{- 9}$ mole ${litre}^{- 1}$

$4 \times 10^{- 8}$ mole ${litre}^{- 1}$

answer is D.

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Detailed Solution

In the following reaction

${Ni}^{2 +} + 6 {NH}_{3} ⇌ {[Ni {({NH}_{3})}_{6}]}^{2 +}; K_{f} = 6 \times 10^{8}$

$\begin{array}{cc} 50 \times 0.2 & 50 \times 0.2 \\ = 10 & = 100 \\ ~ 0 & 40 \end{array}$ $- - 10$

$6 \times 10^{8} = \frac{\frac{10}{100}}{[{Ni}^{2 +}] \times {(\frac{40}{100})}^{6}} \Rightarrow [{Ni}^{2 +}] = 4.07 \times 10^{- 8}$

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