Respuesta :
Answer:
[tex][Ni^{2+}]= 2.327*10^{-8} M[/tex]
[tex][Ni(NH_3)_6]^{2+} _{(aq)}]= 0.3 M[/tex]
Explanation:
From the question we are told that
The number of moles of [tex]Ni(NH_3)_6Cl_2[/tex] is [tex]n = 0.10 \ moles[/tex]
The volume of [tex]NH_3[/tex] is = [tex]V = 0.50L[/tex]
The concentration of [tex]NH_3[/tex] [tex]C = 3.0M[/tex]
The [tex]k_{overall}[/tex] for [tex][Ni(NH_3) _6^{2+}][/tex] is [tex]k_{overall} = 5.5*10^{8}[/tex]
The number of moles of [tex]NH_3[/tex] is mathematically evaluated as
[tex]No\ of mole = concentration \ * \ volume[/tex]
substituting values
[tex]Moles = 0.50* 3.0[/tex]
[tex]=1.5 \ moles[/tex]
The equation for this reaction is
[tex]Ni^{2+}_{(aq)} + 6NH_3_{(aq)}[/tex] ⇄ [tex][Ni(NH_3)_6]^{2+} _{(aq)}[/tex]
From this equation we can mathematically evaluate the limiting reactant as
For [tex]Ni^{2+}[/tex]
[tex]0.10moles [Ni ^{2+}] * \frac{6moles NH_3}{1 mole [N_i^{2+}]}[/tex]
[tex]= 0.6 moles [Ni^{2+}][/tex]
For [tex]NH_3[/tex]
[tex]1.5 moles [NH_3] * \frac{6moles NH_3}{1 mole [N_i^{2+}]}[/tex]
[tex]= 9 moles [NH_3][/tex]
Hence the Limiting reactant is [tex]Ni^{2+}[/tex]
At the initial the number of moles of [tex][Ni(NH_3)_6]^{2+} _{(aq)}[/tex] produced is mathematically evaluated as
[tex]0.10 [Ni^{2+}] *\frac{1 moles [Ni(NH_3)_6]^{2+}}{1 mol [Ni^{2+}]}[/tex]
[tex]= 0.10 \ moles[/tex]
The remaining amount of [tex]NH_3[/tex] is [tex]1.5 - 0.6 = 0.9 \ moles[/tex]
Now the reaction is complete
The number of moles of [tex]Ni^{2+}[/tex] = 0 moles
For this kind of reaction after the complex solution(which is at equilibrium) has been formed they disassociate again ( which is also at equilibrium )
Let the number of moles of [tex]Ni^{2+}[/tex] after disassociation be z
The the number of moles of [tex]NH_3[/tex] after disassociation be 0.9 + 6z
The 6 is from the reaction equation
The the number of moles of [tex][Ni(NH_3)_6]^{2+} _{(aq)}[/tex] after disassociation be [tex]0.10 -z[/tex]
Now the equilibrium constant for this reaction is
[tex]K_{overall} = \frac{[Ni (NH_3) _6]^{2+}}{[Ni^{2+} [NH_3] ^6]}[/tex]
The powers of these concentration is are their moles
substituting value
[tex]5.5*10^{8} = \frac{0.10 - z}{[z] [0.9 + 6z]^6}[/tex]
Since the back reaction is very little so we can neglecting subtracting it from the moles of [tex][Ni(NH_3)_6]^{2+} _{(aq)}[/tex] and adding it to the number of moles of [tex]NH_3[/tex]
Because it won't affect the value of z obtained
[tex]5.5*10^{8} = \frac{0.10}{[z] [0.5]^6}[/tex]
[tex]z = \frac{0.10}{5.5*10^8 * (0.9)}[/tex]
[tex]= 1.16*10^{-8} \ mol[/tex]
So the number of moles of [tex]Ni^{2+}[/tex] at equilibrium is [tex]= 1.16*10^{-8} \ mol[/tex]
The concentration of [tex]Ni^{2+}[/tex] is mathematically represented as
[tex]concentration = \frac{mole}{volume}[/tex]
[tex]= \frac{1.1636*10^{-8}}{0.50}[/tex]
[tex][Ni^{2+}]= 2.327*10^{-8} M[/tex]
The moles of [tex][Ni(NH_3)_6]^{2+} _{(aq)}[/tex] is = [tex]0.10 - 2.327*10^{-8}[/tex]
[tex]= 0.10 \ moles[/tex]
The concentration of [tex][Ni(NH_3)_6]^{2+} _{(aq)}[/tex] is mathematically represented as
[tex]concentration = \frac{mole}{volume}[/tex]
[tex]= \frac{0.10}{0.50}[/tex]
[tex][Ni(NH_3)_6]^{2+} _{(aq)}]= 0.3 M[/tex]
