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The first step in solving any acid-base equilibrium problem is writing the ionization equation. This equation depicts how the acid dissociates in water. For chloroacetic acid (\text{ClCH}_{2} \text{COOH}), the ionization equation is: \[ \text{ClCH}_{2} \text{COOH} \rightarrow \text{ClCH}_{2} \text{COO}^{-} + \text{H}^{+} \] This reaction shows the acid donating a proton (\text{H}^{+}) and forming its conjugate base (\text{ClCH}_{2} \text{COO}^{-}). Understanding the ionization helps in setting up the equilibrium expressions that follow.

The acid dissociation constant, \( K_a \), quantifies the strength of the acid in water. It is derived from the concentrations of the products and the reactants at equilibrium. For chloroacetic acid: \[ K_a = \frac{[\text{ClCH}_{2} \text{COO}^{-}][\text{H}^{+}]}{[\text{ClCH}_{2} \text{COOH}]} \]This equation tells us that the strength of the acid (its tendency to donate protons) is directly related to the concentration of protons and conjugate base produced, and inversely related to the concentration of the undecomposed acid.

To work with acid dissociation constants, we often need to convert \( pK_a \) to \( K_a \). The \( pK_a \) value is the negative logarithm of \( K_a \):\[ pK_a = -\text{log}(K_a) \]Conversely:\[ K_a = 10^{-pK_a} \]In the given problem, chloroacetic acid has a \( pK_a \) of 2.87. Converting this to \( K_a \) gives:\[ K_a = 10^{-2.87} \thickapprox 1.35 \times 10^{-3} \]This conversion is crucial because \( K_a \) is used in the equilibrium expressions to solve for concentrations.

Setting up the initial concentrations and calculating the changes upon reaching equilibrium is key. If the initial concentration of chloroacetic acid is 1.25 M, we assume it ionizes to produce \( x \) M each of \( \text{ClCH}_{2} \text{COO}^{-} \) and \( \text{H}^{+} \). At equilibrium, the concentrations will be:\[ [\text{ClCH}_{2} \text{COOH}] = 1.25 - x \]\[ [\text{ClCH}_{2} \text{COO}^{-}] = x \]\[ [\text{H}^{+}] = x \]Substituting these in the \( K_a \) expression, we get:\[ 1.35 \times 10^{-3} = \frac{x^2}{1.25 - x} \]Assuming \( x \) is small compared to 1.25, we simplify it to:\[ 1.35 \times 10^{-3} \thickapprox \frac{x^2}{1.25} \]Solving for \( x \) provides the equilibrium concentrations of all species.

The pH of a solution is a measure of its acidity, calculated using the concentration of \( \text{H}^{+} \) ions. The formula for pH is: \[ pH = -\text{log}[\text{H}^{+}] \]Given our calculated \( [\text{H}^{+}] = 4.11 \times 10^{-2} \text{M} \), we find:\[ pH = -\text{log}(4.11 \times 10^{-2}) \thickapprox 1.39 \]This pH value indicates the acidity of the chloroacetic acid solution. Alongside, we find the equilibrium concentrations: \[ [\text{ClCH}_{2} \text{COO}^{-}] = 4.11 \times 10^{-2} \text{M} \]and \[ [\text{ClCH}_{2} \text{COOH}] = 1.21 \text{M} \] These calculations provide a comprehensive understanding of the solution's composition at equilibrium.