Problem 67 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The concentration of \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) ion in a solution prepared by adding \(0.1\) mole of \(\mathrm{CH}_{3} \mathrm{COOAg}(\mathrm{s})\) in \(1 \mathrm{~L}\) of \(0.1 \mathrm{M}-\mathrm{HCl}\) solution is [Given: \(K_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=10^{-5}\); \(\left.K_{\mathrm{sp}}(\mathrm{AgCl})=10^{-10} ; K_{\mathrm{sp}}\left(\mathrm{CH}_{3} \mathrm{COOAg}\right)=10^{-8}\right]\) (a) \(10^{-3} \mathrm{M}\) (b) \(10^{-2} \mathrm{M}\) (c) \(10^{-1} \mathrm{M}\) (d) \(1 \mathrm{M}\)

Step-by-Step Solution

Answer

\(10^{-1}\ M\)

Step 1 - Understanding the Reaction

We need to consider the dissolution of \(\mathrm{CH}_{3}\mathrm{COOAg}\text{(s)}\) to its ions and the subsequent reaction with \(\mathrm{HCl}\). The relevant chemical reactions are: \[\mathrm{CH}_{3}\mathrm{COOAg(s)} \xrightarrow{\text{dissolves}} \mathrm{CH}_{3} \mathrm{COO}^{-}(aq) + \mathrm{Ag}^{+}(aq)\] \[\mathrm{HCl(aq)} + \mathrm{Ag}^{+}(aq) \rightarrow \mathrm{AgCl(s)} + \mathrm{H}^{+}(aq)\] Since \(K_{\mathrm{sp}}(\mathrm{AgCl})\) is very low, \(\mathrm{Ag}^{+}\) ions will react completely with \(\mathrm{HCl}\) to form insoluble \(\mathrm{AgCl(s)}\). We can assume that all \(\mathrm{Ag}^{+}\) ions precipitate.

Step 2 - Calculating the Reaction Completion

Initially the concentration of \(\mathrm{HCl}\) is \(0.1\ M\) and we add \(0.1\) mole of \(\mathrm{CH}_{3} \mathrm{COOAg}\text{(s)}\). Since the \(K_{\mathrm{sp}}(\mathrm{AgCl})\) is very small, \(\mathrm{Ag}^{+}\) ions react with \(\mathrm{HCl}\) to completion, removing \(0.1\ mol\) of \(\mathrm{H}^{+}\) ions and producing \(0.1\ mol\) of \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) ions. The final concentration of \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) ions is the amount of moles divided by the volume of the solution, which is \(1\ L\), resulting in \(0.1\ M\) concentration.

Key Concepts

Chemical Equilibrium

In chemistry, chemical equilibrium refers to the state where the concentrations of reactants and products remain constant over time because the rates of the forward and reverse reactions are equal. This doesn't mean that the reactants and products are in equal concentrations, but that their ratios don't change. Understanding equilibrium is fundamental in predicting how chemical reactions occur and to what extent.

• Every reversible chemical reaction reaches a state of equilibrium.
• At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.
• The equilibrium can be disturbed by changing the concentrations of reactants or products, temperature, or pressure, described by Le Châtelier's Principle.
• Calculating the precise position of equilibrium requires the constants such as Ka and Ksp, which are specific for each reaction.

Please remember that the existence of a chemical equilibrium is essential for almost all processes happening in a solution, including the dissolution of salts and the dissociation of acids and bases.

Solubility Product Constant (Ksp)

Moving towards a more specific equilibrium concept, the solubility product constant (Ksp) is crucial for understanding ionic compounds' solubility in solution. It represents the level at which a solute (ionic compound) dissolves in solution to form its constituent ions before reaching a point where no further dissolution is possible, resulting in a saturated solution.

• Ksp is specific to each compound and depends on temperature.
• A higher Ksp value means that more solute can dissolve before reaching saturation.
• When ions from a saturated solution form a solid, the system is at equilibrium, and no net change is occurring in the concentration of ions.
• Predicting if a precipitate will form in a solution can be done by comparing the reaction quotient (Q) to the Ksp value.

In essence, Ksp aids in predicting both the extent to which a salt can dissolve and the conditions under which a precipitate will form.

Acid Dissociation Constant (Ka)

The acid dissociation constant (Ka) is a particular example of an equilibrium constant that measures the strength of an acid in solution. It quantifies the extent to which an acid can donate and hydrogen ions (H+) dissociate in a solution.

• A larger Ka value indicates a stronger acid, meaning it donates H+ ions more fully.
• Weak acids have a smaller Ka, which implies that they do not fully dissociate and equilibrium lies far to the left (favoring the undissociated acid form).
• Understanding Ka is essential for calculating the pH of a solution and for predicting the direction of acid-base reactions.
• pKa, the negative logarithm of Ka, is often used to express acid strength to avoid dealing with very small decimal numbers.

Acids and their dissociation are a cornerstone of chemistry because they interact with bases to form salts and water, and play a pivotal role in buffer solutions, which resist changes in pH.