Respuesta :
Answer: The concentration of [tex]NH_3[/tex] required will be 0.285 M.
Explanation:
To calculate the molarity of [tex]NiC_2O_4[/tex], we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Moles of [tex]NiC_2O_4[/tex] = 0.016 moles
Volume of solution = 1 L
Putting values in above equation, we get:
[tex]\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M[/tex]
For the given chemical equations:
[tex]NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}[/tex]
[tex]Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9[/tex]
Net equation: [tex]NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?[/tex]
To calculate the equilibrium constant, K for above equation, we get:
[tex]K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48[/tex]
The expression for equilibrium constant of above equation is:
[tex]K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}[/tex]
As, [tex]NiC_2O_4[/tex] is a solid, so its activity is taken as 1 and so for [tex]C_2O_4^{2-}[/tex]
We are given:
[tex][[Ni(NH_3)_6]^{2+}]=0.016M[/tex]
Putting values in above equations, we get:
[tex]0.48=\frac{0.016}{[NH_3]^6}}[/tex]
[tex][NH_3]=0.285M[/tex]
Hence, the concentration of [tex]NH_3[/tex] required will be 0.285 M.
