The formation constant of Ni(NH3)62+ is 6 /times 108 at 25°C. If 50 ml of 2.0 M NH3 is added to 50 ml of 0.20 M solution of Ni2+, the concentration of Ni2+ ion will be nearly equal to:

Question

The formation constant of Ni(NH3)62+ is 6 /times 108 at 25°C. If 50 ml of 2.0 M NH3 is added to 50 ml of 0.20 M solution of Ni2+, the concentration of Ni2+ ion will be nearly equal to: (A) 3 ×10-10 mole litre-1 (B) 2 x 10-10 mole litre-1 (C) 2× 10-9 mole litre-1 (D) 4 × 10-8 mole litre-1

Tardigrade Admin · Accepted Answer

Ni2+ + 6NH3 leftharpoons underset0[Ni(NH3)]+6; Kf = 6 × 108 t=0 0.01 mole 0.1 mole Kc =([Ni(NH3)6+6]/[Ni+2][NH3]6) = ((0.1)/[Ni2+](0.4)6) [Ni2+] =6 × 10-8

Q. The formation constant of $N i (N H_{3})_{6}^{2 +}$ is $6/ t im es 1 0^{8}$ at $25° C .$ If $50 m l$ of $2.0 M N H_{3}$ is added to $50 m l$ of $0.20 M$ solution of $N i^{2 +}$ , the concentration of $N i^{2 +}$ ion will be nearly equal to:

$3 \times 1 0^{- 10}$ mole litre $^{- 1}$

$2 x 1 0^{- 10}$ mole litre $^{- 1}$

$2 \times 1 0^{- 9}$ mole litre $^{- 1}$

$4 \times 1 0^{- 8}$ mole litre $^{- 1}$

$N i^{2 +} + 6 N H_{3} ⇌ 0 [N i (N H_{3})]^{+ 6}; K_{f} = 6 \times 1 0^{8}$
$t = 0 0.01$ mole 0.1 mole
$K_{c} = \frac{[ N i ( N H _{3} ) _{6}^{+ 6} ]}{[ N i ^{+ 2} ] [ N H _{3} ] ^{6}}$
$= \frac{( 0.1 )}{[ N i ^{2 +} ] ( 0.4 ) ^{6}}$
$[N i^{2 +}] = 6 \times 1 0^{- 8}$

Q. The formation constant of Ni(NH3​)62+​ is 6/times108 at 25°C. If 50ml of 2.0MNH3​ is added to 50ml of 0.20M solution of Ni2+, the concentration of Ni2+ ion will be nearly equal to:

3×10−10 mole litre−1

2x10−10 mole litre−1

2×10−9 mole litre−1

4×10−8 mole litre−1