Answer:
[tex]1.684*10^{- 9} M[/tex]
Explanation:
Equation for the reaction can be expressed as:
[tex]Zn^{2+}_{(aq)} + 4NH_{3(aq)}[/tex] ⇄ [tex][Zn(NH_3)_4^+_{(aq)}][/tex]
From the equation above; we use the ICE table to determine how much complex ion forms in order to be able to calculate the [tex][Zn^{2+}][/tex] at equilibrium.
So;
[tex]Zn^{2+}_{(aq)}[/tex] [tex]+[/tex] [tex]4NH_{3(aq)}[/tex] ⇄ [tex][Zn(NH_3)_4^+_{(aq)}][/tex]
Initial [tex]1.2*10^{-3}[/tex] [tex]0.130[/tex] [tex]0.0[/tex]
Change [tex]-1.2*10^{-3}[/tex] [tex]-4(1.2*10^{-3})[/tex] [tex]+1.2*10^{-3}[/tex]
Equilibrium [tex]0[/tex] [tex]0.1252[/tex] [tex]1.2*10^{-3}[/tex]
[tex]K_f[/tex] [tex]=[/tex] [tex]\frac{[Zn(NH_3)_4^{2+}]}{[Zn^{2+}][NH_3]^4}[/tex] ⇒ [tex][Zn^{2+}][/tex]
= [tex]\frac{[Zn(NH_3)_4^{2+}]}{[K_fNH_3]^4}[/tex]
= [tex]\frac{(1.2*10^{-3})}{(2.9*10^9)(0.1252)^4}[/tex]
= [tex]1.684*10^{- 9} M[/tex]
Therefore, the concentration of Zn2+(aq) after the solution reaches equilibrium = [tex]1.684*10^{- 9} M[/tex]
The concentration of Zn2+(aq) after reaching equilibrium is mathematically given as
Kf= 1.684*10^{-9} M
Question Parameter(s):
A solution is made that is 1.2×10−3 M in Zn(NO3)2
and 0.130 M in NH3.
Generally, the equation for the Chemical Reaction is mathematically given as
Zn^{2+}_{(aq)}+4NH_{3(aq)} -----><---- [Zn(NH_3)_4^+_{(aq)}]
Therefore
[tex]Kf=\frac{[Zn(NH_3)_4^{2+}]}{[K_fNH_3]^4}[/tex]
[tex]Kf= \frac{(1.2*10^{-3})}{(2.9*10^9)(0.1252)^4}[/tex]
Kf= 1.684*10^{-9} M
In conclusion, the concentration of Zn2+(aq)
Kf= 1.684*10^{-9} M
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