Problem 103 - Chemistry An Atoms First Approach

Question

Calculate the solubility of \(\mathrm{AgCN}(s)\left(K_{\mathrm{sp}}=2.2 \times 10^{-12}\right)\) in a solution containing \(1.0 M \mathrm{H}^{+} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN} \text { is } 6.2 \times 10^{-10} .\right)\)

Step-by-Step Solution

Answer

The solubility of AgCN in the given solution containing 1.0 M H⁺ is \(1.48 \times 10^{-6}\) M.

Step 1: 1. Write the dissolution and acid-base equilibrium reactions.

First, let's write the chemical equilibrium equation for the dissolution of AgCN: \[\mathrm{AgCN}(s) \rightleftharpoons \mathrm{Ag}^{+}(aq) + \mathrm{CN}^{-}(aq)\] Next, we write the chemical equilibrium equation for the acid-base reaction involving H⁺ and CN⁻: \[\mathrm{H}^{+}(aq) + \mathrm{CN}^{-}(aq) \rightleftharpoons \mathrm{HCN}(aq)\]

Step 2: 2. Write the Ksp and Ka expressions.

Now, let's write the equilibrium expressions corresponding to the Ksp and Ka values given. For the dissolution of AgCN: \[K_{\mathrm{sp}} = [\mathrm{Ag}^{+}][\mathrm{CN}^{-}]\] For the acid-base reaction between H⁺ and CN⁻: \[K_{\mathrm{a}} = \frac{[\mathrm{HCN}]}{[\mathrm{H}^{+}][\mathrm{CN}^{-}]}\]

Step 3: 3. Define the terms in the equilibrium expressions.

Let x be the solubility of AgCN in moles per liter. Then at equilibrium: \[[\mathrm{Ag}^{+}] = x\] \[[\mathrm{CN}^{-}] = x\] \[[\mathrm{HCN}] = x\] \[[\mathrm{H}^{+}] = 1.0\,\mathrm{M}\text, since it is given that the solution contains 1.0 M H⁺)

Step 4: 4. Substitute the equilibrium concentrations into the Ksp and Ka expressions.

Now, we substitute the equilibrium concentrations into the Ksp and Ka expressions: For Ksp: \[2.2 \times 10^{-12} = x^2\] For Ka: \[6.2 \times 10^{-10} = \frac{x}{1.0\,\mathrm{M} \times x}\]

Step 5: 5. Solve the equations to find the solubility of AgCN.

First, solve for x using the Ka expression: \[6.2 \times 10^{-10} = \frac{x}{x}\] \[\text{Since the value of HCN is negligible, we have:}\] \[[\mathrm{HCN}] \approx 0\] Now, solve for x using the Ksp expression: \[2.2 \times 10^{-12} = x^2\] \[x = \sqrt{2.2 \times 10^{-12}}\] \[x = \mathrm{1.48 \times 10^{-6}\, M}\] Therefore, the solubility of AgCN in the given solution is \(1.48 \times 10^{-6}\) M.

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, often called the Ksp, is a crucial concept in understanding the solubility of sparingly soluble ionic compounds. It represents the maximum amount of a solid that can dissolve in a solution at a given temperature. When a solid dissolves, it breaks into its respective ions in a reversible reaction. For example, for the dissolution of silver cyanide (AgCN), the reaction is:

• \[\mathrm{AgCN}(s) \rightleftharpoons \mathrm{Ag}^{+}(aq) + \mathrm{CN}^{-}(aq)\]

The Ksp expression for this reaction is given by the concentrations of the resulting ions:

• \[K_{\mathrm{sp}} = [\mathrm{Ag}^{+}][\mathrm{CN}^{-}]\]

The Ksp value helps predict how much of the solid can dissolve to create a saturated solution. A small Ksp value, like \(2.2 \times 10^{-12}\) for AgCN, indicates low solubility. This means that only a small amount of AgCN can dissolve before the solution becomes saturated. Calculating the solubility from Ksp involves setting up an equilibrium expression using stoichiometry and solving for the concentration of ions.

Acid-Base Equilibrium

In a solution where both solubility and acid-base reactions occur, it is important to understand acid-base equilibrium. In the case of the AgCN solution, \(\mathrm{CN}^{-}\) ions can react with \(\mathrm{H}^{+}\) ions. This reaction forms HCN, an example of an acid-base reaction, and is represented by:

• \[\mathrm{H}^{+}(aq) + \mathrm{CN}^{-}(aq) \rightleftharpoons \mathrm{HCN}(aq)\]

The equilibrium constant for this reaction is Ka, given by:

• \[K_a = \frac{[\mathrm{HCN}]}{[\mathrm{H}^{+}][\mathrm{CN}^{-}]}\]

With an initial concentration of \(1.0 \, \mathrm{M}\, \mathrm{H}^{+} \), the presence of the strong acid influences the extent to which the \( \mathrm{CN}^{-} \) ions are available to dissolve AgCN. Solving for the solubility involves considering both the Ksp for solubility and Ka for acid strength. By understanding how \(\mathrm{H}^+\) shifts equilibrium towards forming \(\mathrm{HCN}\), we can calculate effective ion concentrations at equilibrium, illustrating how the acidity of the solution influences solubility.

Equilibrium Expressions

Equilibrium expressions are mathematical representations that describe the balance between reactants and products in a chemical reaction at equilibrium. These expressions are vital for calculating various concentrations in reactions involving solubility and acid-base systems. Each equilibrium expression is derived based upon the balanced equation of the reaction.For reactions like those seen in the solubility of \(\mathrm{AgCN}\) in an acidic solution, we must derive separate expressions for both Ksp and Ka:

• For dissolution: \[K_{\mathrm{sp}} = [\mathrm{Ag}^+][\mathrm{CN}^-]\]
• For acid-base interaction: \[K_{\mathrm{a}} = \frac{[\mathrm{HCN}]}{[\mathrm{H}^{+}][\mathrm{CN}^-]}\]

Using variables to denote the equilibrium concentrations, these expressions provide the foundation for calculating unknowns within the system. By substituting known values into the expressions, such as initial hydrogen ion concentrations and total equilibrium concentrations, we can solve for the solubility of the ionic compound. Understanding these principles allows students to explore how complex interactions in chemistry affect the observed properties of solutions.