What is the concentration of Ni2+ in 0.0775M Ni(NO3)2 and 1.3981M NH3
5.0*10^8= 0.0775/0.9331^6x
x= 2.35e-10
To determine the concentration of Ni2+ in the solution of Ni(NO3)2 and NH3, we need to consider the chemical reaction that occurs between Ni(NO3)2 and NH3, which forms a complex ion.
The balanced equation for the reaction is:
Ni(NO3)2 + 6NH3 → [Ni(NH3)6]2+ + 2NO3-
From the equation, we can see that every one mole of Ni(NO3)2 reacts with 6 moles of NH3 to form one mole of [Ni(NH3)6]2+. Therefore, the molar ratio between [Ni(NH3)6]2+ and Ni(NO3)2 is 1:1.
Given:
- Concentration of Ni(NO3)2 solution = 0.0775 M
- Concentration of NH3 solution = 1.3981 M
To determine the concentration of Ni2+ in the solution, we need to find the initial moles of Ni(NO3)2 reacted and then calculate the concentration of Ni2+.
1. Determine the initial moles of Ni(NO3)2:
Moles of Ni(NO3)2 = concentration × volume
Moles of Ni(NO3)2 = 0.0775 M × volume (in liters)
2. Moles of [Ni(NH3)6]2+ formed:
From the balanced equation, the moles of [Ni(NH3)6]2+ formed will be the same as the moles of Ni(NO3)2 used, as they have a 1:1 molar ratio.
3. Concentration of Ni2+:
Since the molar ratio between [Ni(NH3)6]2+ and Ni2+ is 1:1, the concentration of Ni2+ will also be 0.0775 M.
So, the concentration of Ni2+ in the given solution of Ni(NO3)2 and NH3 is 0.0775 M.