To determine the concentration of [tex]\( \text{Ni}^{2+}(aq) \)[/tex] ion in a [tex]\( 0.035 \, M \, \text{Ni}(NO_3)_2 \)[/tex] solution that is also [tex]\( 1.00 \, M \, \text{NH}_3 \)[/tex], we need to use the given formation constant [tex]\( K_f \)[/tex] for [tex]\( \text{Ni(NH}_3)_6^{2+} \)[/tex]. The formation constant [tex]\( K_f \)[/tex] expresses the equilibrium concentration of the complex ion formed from the metal ion and the ligands.
The given data is:
- [tex]\( \text{Ni(NO}_3)_2 \, \text{concentration} = 0.035 \, M \)[/tex]
- [tex]\( \text{NH}_3 \, \text{concentration} = 1.00 \, M \)[/tex]
- [tex]\( K_f = 5.5 \times 10^8 \)[/tex]
The complex ion formation reaction is:
[tex]\[ \text{Ni}^{2+} + 6 \text{NH}_3 \rightleftharpoons \text{Ni(NH}_3\text{)}_6^{2+} \][/tex]
At equilibrium, we can write an expression for [tex]\( K_f \)[/tex]:
[tex]\[ K_f = \frac{[\text{Ni(NH}_3\text{)}_6^{2+}]}{[\text{Ni}^{2+}] [\text{NH}_3]^6} \][/tex]
To find the concentration of [tex]\( \text{Ni}^{2+}(aq) \)[/tex], re-arrange the equation to solve for [tex]\( [\text{Ni}^{2+}] \)[/tex]:
[tex]\[ [\text{Ni}^{2+}] = \frac{[\text{Ni(NO}_3)_2]}{K_f [\text{NH}_3]^6} \][/tex]
Substitute the given values into the above equation:
[tex]\[ [\text{Ni}^{2+}] = \frac{0.035}{5.5 \times 10^8 \times (1.00)^6} \][/tex]
Simplifying the calculation:
[tex]\[ [\text{Ni}^{2+}] = \frac{0.035}{5.5 \times 10^8} \][/tex]
Simplify further:
[tex]\[ [\text{Ni}^{2+}] = \frac{0.035}{550000000} \][/tex]
Perform the division:
[tex]\[ [\text{Ni}^{2+}] = 6.363636363636364 \times 10^{-11} \, M \][/tex]
Thus, the concentration of [tex]\( \text{Ni}^{2+}(aq) \)[/tex] is approximately [tex]\( 6.4 \times 10^{-11} \, M \)[/tex].
The correct answer among the given options is:
[tex]\[ 6.4 \times 10^{-11} \, M \][/tex]
