Consider the insoluble compound cobalt(II) carbonate , CoCO3 . The cobalt(II) ion also forms a complex with ammonia . Write a balanced net ionic equation to show why the solubility of CoCO3 (s) increases in the presence of ammonia and calculate the equilibrium constant for this reaction. For Co(NH3)62+ , Kf = 7.7×104 . Use the pull-down boxes to specify states such as (aq) or (s).

7.7×10^-6

Explanation:

CoCO3(s)<> Co^2+(aq) + CO3^2-(aq) Ksp= 1.0 ×10 ^-10

Co^2+(aq) + 6NH3(aq) <> [Co(NH3)6]^2+. Kf = 7.7×10^4

Overall:

CoCO3(s) + 6NH3(aq) <> [Co(NH3)6]^2+(aq) + CO3^2-(aq) Knet= Ksp× Kf

Knet= Ksp× Kf = 1.0 ×10 ^-10 ×7.7×10^4= 7.7×10^-6

The insoluble CoCO3 dissolves in the presence of ammonia because bof the formation of hexamine cobalt II complex ion.

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).