Question

A 0.130 mole quantity of NiCl 2 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Ni 2 + ions at equilibrium? Assume the formation constant of Ni ( NH 3 ) 2 + 6 is 5.5 × 10

Answer 1

This is an incomplete question, here is a complete question.

A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸

Answer : The concentration of
`Ni^(2+)` ions at equilibrium is,
`4.31* 10^(-8)`

Explanation : Given,

Moles of
`NiCl_2` = 0.130 mol

Volume of solution = 1 L

`Concentration=(Moles )/(Volume)`

Concentration of
`NiCl_2` = Concentration of
`Ni^(2+)` = 0.130 M

Concentration of
`NH_3` = 1.20 M

`K_f=5.5* 10^8`

The equilibrium reaction will be:

`Ni^(2+)(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^(2+)`

Initial conc. 0.130 1.20 0

At eqm. x [1.20-6(0.130)] 0.130

= 0.42

The expression for equilibrium constant is:

`K_f=([Ni(NH_3)_6^(2+)])/([Ni^(2+)][NH_3]^6)`

Now put all the given values in this expression, we get:

`5.5* 10^8=((0.130))/((x)* (0.42)^6)`

`x=4.31* 10^(-8)`

Thus, the concentration of
`Ni^(2+)` ions at equilibrium is,
`4.31* 10^(-8)`