A. In a reaction of Ni2+ ions with ammonia, the only complex ion to form is the hexaammine species, [Ni(NH3)6]2+. Given a solution where the initial [Ni2+] is 0.2M, and the initial [NH3] is 4M and that  β6= 6 x 108, calculate the equilibrium concentration of the Ni2+ ion. ANS: The concentration of Ni2+ ions is

To solve this problem, we need to use the formula for the equilibrium constant for the formation of the complex ion, which is given by:

β6 = [Ni(NH3)6]2+ / ([Ni2+] [NH3]^6)

We are asked to find the equilibrium concentration of Ni2+, so let's denote it as [Ni2+]eq. At equilibrium, the concentration of the hexaammine complex will be the initial concentration of Ni2+ minus [Ni2+]eq, or 0.2 - [Ni2+]eq. The concentration of NH3 at equilibrium will be the initial concentration minus six times the concentration of the hexaammine complex, or 4 - 6*(0.2 - [Ni2+]eq).

Substituting these into the formula gives:

β6 = (0.2 - [Ni2+]eq) / ([Ni2+]eq * (4 - 6*(0.2 - [Ni2+]eq))^6)

We know that β6 = 6 x 10^8, so we can solve this equation for [Ni2+]eq. This is a complex equation that may require numerical methods to solve. However, if the formation of the complex is nearly complete, we can make the approximation that [Ni2+]eq is much smaller than 0.2 M, and that the change in the concentration of NH3 is negligible. This simplifies the equation to:

6 x 10^8 = 0.2 / ([Ni2+]eq * 4^6)

Solving this for [Ni2+]eq gives the equilibrium concentration of Ni2+.