Question

Will 0.10 mol of AgBr completely dissolve in 1.0 \(\mathrm{L}\) of 3.0 \(\mathrm{M} \mathrm{NH}_{3} ?\) The \(K_{\mathrm{sp}}\) value for \(\mathrm{AgBr}(s)\) is \(5.0 \times 10^{-13},\) and the overall formation constant for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is \(1.7 \times 10^{7}\) , that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$

Short answer

The solubility of AgBr in the presence of 3.0 M NH₃ is calculated to be approximately \(7.1 \times 10^{-7}\ \mathrm{M}\). Given the maximum amount of AgBr that can dissolve is about \(7.1 \times 10^{-7}\ \mathrm{mol}\) in 1.0 L of NH₃ solution, which is much smaller than the given amount of AgBr (0.10 mol), it will not completely dissolve.

Step-by-step solution

Write the Ksp expression for AgBr

The solubility product (Ksp) expression for AgBr can be written as follows: \[K_{sp} = [\mathrm{Ag}^+][\mathrm{Br}^-]\] We are given the value of Ksp as \(5.0 \times 10^{-13}\).

Write the complex formation reaction and its equilibrium expression

The formation of the complex ion Ag(NH3)2+ can be represented as: \[\mathrm{Ag}^+ + 2 \mathrm{NH}_3 \rightleftharpoons \mathrm{Ag}(\mathrm{NH}_3)_2^+\] The equilibrium expression for this reaction is: \[K = \frac{[\mathrm{Ag}(\mathrm{NH}_3)_2^+]}{[\mathrm{Ag}^+][\mathrm{NH}_3]^2}\] We are given the value of K as \(1.7 \times 10^{7}\).

Relate the equilibrium expressions with the solubility of AgBr

Let the solubility of AgBr in moles per liter (molar solubility) be represented by S. The concentration of Ag+ will be equal to S, and the concentration of Br- will also be equal to S. The concentration of NH3 is given as 3.0 M. Using the equilibrium expression for the complex formation, we can write: \[1.7 \times 10^7 = \frac{[\mathrm{Ag}(\mathrm{NH}_3)_2^+]}{[\mathrm{Ag}^+][\mathrm{NH}_3]^2}\] We can relate this to the solubility of AgBr: \[1.7 \times 10^7 = \frac{[\mathrm{Ag}(\mathrm{NH}_3)_2^+]}{(S)(3)^2}\] Solve for the concentration of Ag(NH3)2+: \[[\mathrm{Ag}(\mathrm{NH}_3)_2^+] = (1.7 \times 10^7)(S)(3)^2\]

Use the Ksp expression to find the solubility of AgBr

Using the Ksp expression and the relationship in Step 3, we can write: \(5.0 \times 10^{-13} = S (S + (1.7 \times 10^7)(S)(3)^2)\) Simplify the equation and solve for S: \(5.0 \times 10^{-13} = S^2 + (1.53 \times 10^8)S^3\) Since the Ksp value is very small, we can make the approximation that S^3 is much smaller compared to the S^2 term. Therefore, we can ignore the S^3 term and solve for S: \(5.0 \times 10^{-13} \approx S^2\) Solving for S, we get: \[S = \sqrt{5.0 \times 10^{-13}} \approx 7.1 \times 10^{-7}\ \mathrm{M}\]

Calculate the maximum amount of AgBr that can dissolve in 1.0 L solution

Now that we have the solubility of AgBr, we can calculate the maximum amount of AgBr that can dissolve in 1.0 L of the NH3 solution: \[7.1 \times 10^{-7}\ \mathrm{M} \times 1.0\ \mathrm{L} \approx 7.1 \times 10^{-7}\ \mathrm{mol}\]

Compare the maximum amount with the given amount of AgBr

We found that the maximum amount of AgBr that can dissolve in 1.0 L of 3.0 M NH3 solution is approximately 7.1 x 10^-7 mol. Since this value is much smaller than the given amount of AgBr (0.10 mol), the given amount of AgBr will not completely dissolve in the solution.

Key concepts

Complex Ion Formation

Complex ion formation refers to the process where simple ions join with ligands to form complex ions. In the context of Silver Bromide (AgBr) dissolving in ammonia (NH3) solution, the silver ion (Ag⁺) interacts with ammonia to form a more stable complex ion known as \[\mathrm{Ag(NH_3)_2}^+\]This complex ion is quite stable owing to its significant formation constant. The reaction can be written as:

• \( \mathrm{Ag}^+ + 2 \mathrm{NH}_3 \rightleftharpoons \mathrm{Ag(NH_3)_2}^+ \)
• ability of the silver ion to form a complex with ligands like ammonia helps in increasing the apparent solubility.

The stability of this complex assists in further dissolving the AgBr by pulling equilibrium to produce more silver ions, playing a critical role in solubility equilibria.

Solubility Product Constant (Ksp)

The solubility product constant, or Ksp, measures how much of a compound can dissolve in solution. For Silver Bromide (AgBr), the Ksp is very low, i.e.,\(K_{sp} = 5.0 \times 10^{-13}\)This indicates that AgBr has low solubility in pure water. The Ksp expression can be formed by setting the concentrations of the dissolved ions in the product of the equilibrium:\[K_{sp} = [\mathrm{Ag}^+][\mathrm{Br}^-]\]This value is used to calculate how much of each ion is present in a saturated solution of AgBr. It acts as a benchmark for understanding how the presence of other substances like NH3 increases solubility.

Formation Constant

The formation constant represents how likely a complex ion is to form in solution. Also known as the stability constant, it provides a quantitative measure of the equilibrium concentration of complex ions formed from its components. For the complex ion \(\mathrm{Ag(NH_3)_2}^+\), the formation constant (\(K_f\)) is given as:\(K_f = 1.7 \times 10^7\)Such a value signifies the high propensity of silver ions to pair with NH3 to form the complex. It significantly indicates how much the silver ion becomes more soluble when association with NH3 occurs. The equation handling this formation is presented as:\[K = \frac{[\mathrm{Ag(NH_3)_2}^+]}{[\mathrm{Ag}^+][\mathrm{NH}_3]^2}\]A large \(K_f\) suggests that the formation of the complex ion is strongly favored.

Silver Bromide (AgBr)

Silver Bromide (AgBr) is a sparingly soluble salt, which means that it dissolves in water only slightly. Its solubility is influenced typically by both its inherent solubility product constant and external interactions with compounds like NH3. In a scenario where AgBr is exposed to excess ammonia, the formation of complex ions as explained would effectively lower the concentration of silver ions in solution. Under such a setup, the equilibrium shifts, enhancing the solubility of AgBr beyond what's achievable in a pure solvent. The overall reactions including the direct dissolution of AgBr and formation of the complex ion dynamically balance according to the conditions of the mixture. AgBr’s behaviour in various solutions is fundamentally tied to understanding solubility equilibria and ionic interactions.

Equilibrium Expressions

Equilibrium expressions are essential in representing the concentrations of reactants and products in a chemical equilibrium state. These expressions allow chemists to calculate various concentrations and contextualize results driven by equilibrium constants like Ksp and formation constants.Given

• the Ksp expression for AgBr: \( K_{sp} = [\mathrm{Ag}^+][\mathrm{Br}^-]\)
• the formation constant: \( K = \frac{[\mathrm{Ag(NH_3)_2}^+]}{[\mathrm{Ag}^+][\mathrm{NH_3}]^2} \)

We recognize that each equilibrium expression points to a balance between ions. These equations help in predicting whether a precipitate will form or if a complex can increase solubility. By solving such equations, we can determine the maximum concentrations allowable under given conditions. Integrating these equilibria ensures that the theoretical approach aligns well with practical scenarios in chemical systems.