Question

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

• A. $3 \times10^{-10}$ mole litre$^{-1}$
• B. $2 x 10^{-10}$ mole litre$^{-1}$
• C. $2\times 10^{-9}$ mole litre$^{-1}$
• D. $4 \times 10^{-8}$ mole litre$^{-1}$

Answer 1

Answer: (D)

$Ni^{2+} + 6NH_{3} \rightleftharpoons \underset{0}{[Ni(NH_{3})]^{+6}}; K_{f} = 6 \times 10^{8}$
$t=0 \,\,\, 0.01$ mole 0.1 mole
$K_{c} =\frac{[Ni(NH_{3})_{6}^{+6}]}{[Ni^{+2}][NH_{3}]^{6}}$
$= \frac{(0.1)}{[Ni^{2+}](0.4)^{6}}$
$[Ni^{2+}] =6 \times 10^{-8}$