Question

An equilibrium solution of the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) contains \(0.30 \mathrm{M}\) of \(\mathrm{NH}_{4}^{+}\) and \(0.15 \mathrm{M}\) of the actual \(^{+}\) complex ion. To obtain a concentration of \(\mathrm{Ag}^{+}\) equal to \(1.0 \times 10^{-6} \mathrm{M}\), what must the \(\mathrm{pH}\) of this solution be? \(\mathrm{K}_{\text {diss }}\) of \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}=6.0\) \(\times 10^{-8}, \mathrm{~K}_{\mathrm{b}}\) of \(\mathrm{NH}_{3}=1.8 \times 10^{-5}\) and \(\mathrm{K}_{\mathrm{W}}=1 \times 10^{-14}\)

Short answer

The pH of the solution must be approximately 11.08 to obtain a concentration of $\mathrm{Ag}^{+}$ equal to \(1.0 \times 10^{-6} \mathrm{M}\).

Step-by-step solution

Write the dissociation and ionization equations.

Write the equation for the dissociation of Ag(NH3)2+ and the ionization of NH3: \[Ag(NH3)2^+ \rightleftharpoons Ag^+ +2NH3\] \[NH3 + H2O \rightleftharpoons NH4^+ + OH^-\]

Set up the equilibrium expression.

Using the given Kdiss, we can set up the equilibrium expression for the dissociation of Ag(NH3)2+: \[K_{diss} = \frac{[Ag^+][NH3]^2}{[Ag(NH3)2^+]}\]

Relate the equilibrium concentrations.

Let x be the equilibrium concentration of Ag+ when pH is adjusted. Then the equilibrium concentration of NH3 is 2x: \[[Ag^+] = x\] \[[NH3] = 2x\]

Substitute the given values into the equilibrium expression.

Substitute the given Kdiss and equilibrium concentrations into the equation: \[6.0 \times 10^{-8} = \frac{x(2x)^2}{0.15}\]

Solve for x.

Solve the equation for x to find the equilibrium concentration of Ag+: \[x = 1.0 \times 10^{-6}\]

Set up the ionization expression.

Using Kb, set up the ionization expression for NH3: \[K_b = \frac{[NH4^+][OH^-]}{[NH3]}\]

Substitute the given values into the ionization expression.

Substitute the given Kb and equilibrium concentrations into the equation: \[1.8 \times 10^{-5} = \frac{0.30 \times [OH^-]}{2\times (1.0\times 10^{-6})}\]

Solve for [OH-].

Solve the equation for the equilibrium concentration of OH-: \[[OH^-] = 1.2 \times 10^{-3}\]

Find pOH.

Calculate the pOH of the solution using the equilibrium concentration of OH-: \[pOH = -\log(1.2 \times 10^{-3}) = 2.92\]

Find pH.

Use the relationship between pH and pOH to find the pH of the solution: \[pH = 14 - pOH = 14 - 2.92 = 11.08\] The pH of the solution must be approximately 11.08.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a reversible chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentrations of the reactants and products over time. It’s vital to understand that this doesn't mean the reactants and products are equal in concentration, but rather that their ratio remains constant. In the context of complex ions like \(Ag(NH_3)_2^+\), we are looking at a dynamic equilibrium between the metal complex and its constituents – silver ions \(Ag^+\) and ammonia molecules \(NH_3\). At equilibrium, these species exist in concentrations that follow a constant, known as the dissociation constant \(K_{diss}\), which quantifies the propensity of the complex ion to dissociate in solution.

Dissociation Constant

The dissociation constant \(K_{diss}\) is essential for understanding the stability of complex ions in solution. It's a specialized form of the equilibrium constant that applies to the dissociation of a complex ion into its components. In mathematical terms, for a complex ion \(A(B)_n^+\) dissociating into its components \(A^+\) and \(B\), the dissociation constant is defined as \(K_{diss} = \frac{[A^+][B]^n}{[A(B)_n^+]}\). A lower value of \(K_{diss}\) indicates a more stable complex ion, which means it is less likely to split into its ions. Conversely, a higher dissociation constant suggests a complex ion that easily dissociates. When calculating the pH required to achieve a particular concentration of \(Ag^+\) from the complex ion \(Ag(NH_3)_2^+\), the dissociation constant is used to relate the concentrations of the products and reactant at equilibrium.

Ionization Expression

The ionization expression is analogous to the equilibrium expression but is specific to the ionization of a compound in water. For a base like ammonia \(NH_3\), which ionizes in water according to the equation \(NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\), the ionization constant \(K_b\) is used to describe the degree to which the base ionizes. The ionization expression for this reaction is \(K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\). This shows how the concentrations of the ions produced—ammonium \(NH_4^+\) and hydroxide \(OH^-\)—are related to the concentration of the ammonia itself. Understanding this relationship is crucial for adjusting the pH of a solution because it tells us how the concentration of \(OH^-\) ions, which affect pH, will change with varying amounts of ammonia.

pH Calculation

The pH of a solution is a numerical measure of its acidity or basicity. It's calculated as the negative logarithm to the base 10 of the concentration of hydrogen ions in the solution: \(pH = -log[H^+]\). In a situation involving a base, such as ammonia \(NH_3\), we often calculate the pOH instead, which is related to the concentration of hydroxide ions \(OH^-\). The pOH is then calculated similarly: \(pOH = -log[OH^-]\). The pH and pOH are connected through the equation \(pH + pOH = 14\), which comes from the ion product of water \(K_w = [H^+][OH^-] = 1 \times 10^{-14}\) at 25°C. By manipulating these expressions, chemists can control and predict the pH necessary to form or dissociate complex ions in solution, which has practical applications ranging from mining to medicine.