In: Chemistry
You mix a 116.0 mL sample of a solution that is 0.0102 M in NiCl2 with a 165.0 mL sample of a solution that is 0.232 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)
no of moles of Ni^2+ = molarity * volume in L
= 0.0102*0.116 = 0.0011832moles
no of moles of NH3 = molarity * volume in L
= 0.232*0.165 = 0.03828 moles
Ni^2+ (aq) + 6NH3 --------------> [Ni(NH3)6]^2+
I 0.0011832 0.03828 0
C -0.0011832 -6*0.0011832 0.0011832
E 0 0.0311808 0.0011832
x
Kf = [Ni(NH3)6]^2+]/[Ni^2+][NH3]^6
2*10^8 = 0.0011832/[Ni^2+](0.0311808)^6
[Ni^2+] = 0.0011832/2*10^8(0.0311808)^6
[NI^2+] = 0.00643M