Question

Wrige the net ionic equation for the reaction that occurs when ammonia is added to a ni(no3)2 solution?

What is the concentration of Ni2+ (aq) ion in 0.045 M Ni(NO3)2 solution that is also 1 M NH3 (k for Ni(NH3)6 = 5.5E5?

Answer 1

Molecular equation:

Ni(NO3)2 (aq.) + 6 NH3 (aq.) ---------------> [Ni(NH3)6](NO3)2 (aq.)

Ionic equation:

Ni²⁺ (aq.) + 2 NO₃^- (aq.) + 6 NH₃ (aq.) ------> [Ni(NH₃)₆]²⁺ (aq.) + 2 NO₃^- (aq.)

Net ionic equation:

Ni²⁺ (aq.) + 6 NH₃ (aq.) --------> [Ni(NH₃)₆]²⁺ (aq.)

K = [Ni(NH₃)₆]²⁺ / [Ni2+][NH3]⁶

5.5 * 10⁵ = x / (0.045 - x) (1 - 6x)⁶

Since K value too high, 0.045 -x is approx. = x and 1-6x = 6x

5.5 * 10⁵ = x / (x) (6x)⁶

2.57 * 1010 x⁷ = x

x⁶ = 1/ (2.57*10¹⁰)

x = 0.0184 M

Therfore,

at equilibrium, [Ni²⁺] = 0.045 - 0.0184 = 0.027 M