Respuesta :
Explanation:
The given reaction will be as follows.
[tex]Ni^{2+} + 6NH_{3} \rightarrow Ni(NH_{3})^{2+}_{6} + 6H_{2}O \rightarrow Ni(H_{2}O)^{2+}_{6} + 6NH_{3}[/tex]
As [tex]NO^{2-}_{3}[/tex] are the separator ions. Hence, molarity of [tex](NO_{3})_{2}[/tex] will be calculated as follows.
Molarity of [tex](NO_{3})_{2}[/tex] = [tex]\frac{no. of moles of Ni^{2+}}{volume in liter}[/tex]
= [tex]\frac{0.001}{1}[/tex] = 0.001 M
Molarity of [tex]NH_{3}[/tex] = [tex]\frac{no. of moles of NH_{3}}{volume in liter}[/tex]
= [tex]\frac{0.5}{1}[/tex] = 0.5 M
Hence, the molar ratio of [tex]Ni(NO_{3})_{2}[/tex] and [tex]NH_{3}[/tex] is [tex]\frac{1}{6}[/tex].
So, it means that 0.001 moles of [tex](NO_{3})_{2}[/tex] will require 0.001 × 6 = 0.006 moles of [tex]NH_{3}[/tex].
But in actual we are given that there are 0.5 mol of [tex]NH_{3}[/tex].
As molar ratio of [tex](NO_{3})_{2}[/tex] and [tex]Ni(H_{2}O)^{2+}_{6}[/tex] is 1:1.
Therefore, concentration of [tex]Ni(H_{2}O)^{2+}_{6}[/tex] will be calculated as follows.
Concentration of [tex]Ni(H_{2}O)^{2+}_{6}[/tex] = [tex]\frac{/text{no. of moles of [tex]Ni(H_{2}O)^{2+}_{6}}[/tex]}{/text{volume in liter}}[/tex]
= [tex]\frac{0.001}{1}[/tex]
= 0.001 M
Thus, we can conclude that the concentration of [tex]Ni(H_{2}O)^{2+}_{6}[/tex] ions in the solution at equilibrium is 0.001 M.
