Question

From the value of $K_f$ listed in Table $17.1,$ calculate the concentration of ${\mathrm{Ni}}^{2+}$ in $1.0\mathrm{L}$ of a solution that contains a total of $1\times {10}^{-3}\mathrm{mol}$ of nickel(II) ion and that is $0.20\mathrm{M}$ in ${\mathrm{NH}}_3$

Short answer

The concentration of ${\mathrm{Ni}}^{2+}$ in the solution can be determined using the $K_f$ value from Table $17.1$, the equilibrium constant expression, and the initial concentrations of the reactants. The final concentration of ${\mathrm{Ni}}^{2+}$ ions is found by subtracting the amount of nickel ions that reacted with ammonia (x) from the initial amount of nickel ions: $[{\mathrm{Ni}}^{2+}]=1\times {10}^{-3}-x$.

Step-by-step solution

Write out the equilibrium reaction involving ${\mathrm{Ni}}^{2+}$ and ${\mathrm{NH}}_3$

The first step is to write out the balanced chemical equation for the formation of the complex ion between ${\mathrm{Ni}}^{2+}$ and ${\mathrm{NH}}_3$ ligands. For this case, it is given as: $${\mathrm{Ni}}^{2+}(aq)+6{\mathrm{NH}}_3(aq)\rightleftharpoons \mathrm{Ni}({\mathrm{NH}}_3{)}_6^{2+}(aq)$$

Write the equilibrium expression for the formation constant ($K_f$)

The equilibrium expression for this reaction can be written down based on the stoichiometry of the balanced chemical equation as follows: $$K_f=\frac{[\mathrm{Ni}({\mathrm{NH}}_3{)}_6^{2+}]}{[{\mathrm{Ni}}^{2+}][{\mathrm{NH}}_3{]}^6}$$

Establish the concentration of the reactants and products

We know that the total concentration of nickel(II) ion is $1\times {10}^{-3}\mathrm{M}$ and the ammonia concentration is $0.20\mathrm{M}$. We can denote the change in concentration of the species involved in the reaction with variables. Let 'x' be the concentration of $\mathrm{Ni}({\mathrm{NH}}_3{)}_6^{2+}$ formed: $[N{i}^{2+}]=1\times {10}^{-3}-x$ $[N{H}_3]=0.20-6x$ $[Ni(N{H}_3{)}_6^{2+}]=x$

Substitute the concentrations into the equilibrium expression

Now, replace the concentrations in the equilibrium constant expression with the corresponding expressions established in the previous step: $$K_f=\frac{x}{(1\times {10}^{-3}-x)(0.20-6x{)}^6}$$

Determine the value of $K_f$ from Table $17.1$ and solve for x

Using the value of $K_f$ from Table $17.1$ (which needs to be provided or looked up), plug it into the equilibrium expression and solve for x, the concentration of $\mathrm{Ni}({\mathrm{NH}}_3{)}_6^{2+}$: $$K_f=\frac{x}{(1\times {10}^{-3}-x)(0.20-6x{)}^6}$$ Since the $K_f$ value is very large and assuming that $x$ is very small compared to $1\times {10}^{-3}$, we can approximate it as follows: $$K_f\approx \frac{x}{(1\times {10}^{-3})(0.20{)}^6}$$ Solve for x: $$x=K_f\times (1\times {10}^{-3})(0.20{)}^6$$

Determine the final concentration of ${\mathrm{Ni}}^{2+}$

Finally, to determine the concentration of ${\mathrm{Ni}}^{2+}$ ions in the solution, we subtract the amount that has reacted with ammonia (x) from the initial amount of nickel ions: $[{\mathrm{Ni}}^{2+}]=1\times {10}^{-3}-x\text{]}Calculatethefinalconcentrationof\text{(}{\mathrm{Ni}}^{2+}$ by plugging in the value of 'x' obtained in the previous step.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is essential when dealing with reactions that can occur in both forward and reverse directions. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, leading to constant concentrations of the reactants and products, despite both reactions still occurring. This dynamic state can be disrupted by changing conditions such as concentration, temperature, or pressure, following Le Chatelier's principle.

For example, in the context of complex ion formation, adding more ligands or metal ions can shift the equilibrium towards the product side, resulting in more complex ion formation. Conversely, removing a reactant or product will shift the equilibrium to oppose the change. It's crucial to note that equilibrium does not imply equal concentrations of reactants and products but rather a constant ratio as defined by the equilibrium constant.

Formation Constant (Kf)

The formation constant, or stability constant, denoted as K_f, quantifies the stability of a complex ion in a solution. It's a special type of equilibrium constant referring specifically to the formation of complex ions. A high K_f value indicates a stable complex ion that forms readily from its constituent ions.

For students, understanding how to use K_f is paramount in predicting whether a complex will form under certain conditions. In the case of the Ni²⁺ and NH₃ reaction, the high value of K_f allows assumptions that simplify the mathematical approach to finding equilibrium concentrations—assuming that nearly all metal ions react to form the complex, thus simplifying the algebra involved in solving for 'x'.

Equilibrium Expression

An equilibrium expression mathematically represents the ratio of concentration terms for products raised to the power of their stoichiometric coefficients to the reactants raised to the power of their coefficients at equilibrium. The expression for a reaction such as the formation of a complex ion is given by
$$K=\frac{[\text{Products}]}{[\text{Reactants}{]}^{\text{stoichiometric coefficients}}}$$
For students tackling equilibrium problems, it is crucial to write the correct equilibrium expression and understand that only species in aqueous solution or in the gaseous state are included; solids and pure liquids do not appear in this expression.

Using the correct stoichiometry as seen in the given Ni²⁺ and NH₃ reaction is vital for the expression, highlighting that the coefficients become the exponents in the expression. It underscores the balance of chemical equations before using them to write the equilibrium expression.