Problem 83 - Chemistry and Chemical Reactivity

Question

A monoprotic acid HX has \(K_{\mathrm{a}}=1.3 \times 10^{-3} .\) Calculate the equilibrium concentrations of \(\mathrm{HX}\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\) and the \(\mathrm{pH}\) for a \(0.010 \mathrm{M}\) solution of the acid.

Step-by-Step Solution

Answer

Equilibrium [HX] = 0.0064 M, [H⁺] = 0.0036 M, pH = 2.44.

Step 1: Write the Equilibrium Expression

For the dissociation of a monoprotic acid \( HX \) in water, the chemical equation is \( HX \rightleftharpoons H^+ + X^- \). The expression for the equilibrium constant \( K_a \) is given by:\[ K_a = \frac{[H^+][X^-]}{[HX]} \] Where \( K_a = 1.3 \times 10^{-3} \) and the initial concentration of \( [HX] = 0.010 M \).

Step 2: Identify the Initial and Equilibrium Concentrations

Initially, the concentration of \( [HX] \) is \( 0.010 M \), and concentrations of \( [H^+] \) and \( [X^-] \) are 0. At equilibrium, let \( x \) be the change in concentration of \( [H^+] \) and \( [X^-] \), so:- \( [HX] = 0.010 - x \)- \( [H^+] = x \)- \( [X^-] = x \)

Step 3: Substitute into the Equilibrium Expression

Substitute the equilibrium concentrations into the equilibrium expression:\[ K_a = \frac{x \cdot x}{0.010 - x} = \frac{x^2}{0.010 - x} \] Given \( K_a = 1.3 \times 10^{-3} \), the equation becomes:\[ 1.3 \times 10^{-3} = \frac{x^2}{0.010 - x} \]

Step 4: Approximate and Solve the Quadratic Equation

Assume that \( x \) is small compared to 0.010, so \( 0.010 - x \approx 0.010 \). Thus, the equation simplifies to:\[1.3 \times 10^{-3} = \frac{x^2}{0.010} \] Canceling 0.010 from both sides gives:\[x^2 = 1.3 \times 10^{-5} \]Solve for \( x \):\[x = \sqrt{1.3 \times 10^{-5}} \approx 0.0036 \]

Step 5: Calculate Equilibrium Concentrations and pH

Given \( x = 0.0036 \), the equilibrium concentrations are:- \( [H^+] = 0.0036 \) M- \( [HX] = 0.010 - 0.0036 = 0.0064 \) MTo find the pH:\[ \text{pH} = -\log[H^+] = -\log(0.0036) \approx 2.44\]

Key Concepts

Equilibrium Constant (Ka)

The equilibrium constant, known as the acid dissociation constant \(K_a\), is a fundamental concept in chemistry. It tells us how strong a monoprotic acid is by measuring how well it dissociates in solution. For a given acid \(HX\), the equilibrium expression is
\[K_a = \frac{[H^+][X^-]}{[HX]}\]The value of \(K_a\) is specific to each acid and indicates how much the acid will ionize in water. A large \(K_a\) means the acid dissociates well and is strong, while a small \(K_a\) indicates a weaker acid.

• If \(K_a\) is high, the equilibrium lies to the right, favoring products \([H^+][X^-]\).
• If \(K_a\) is low, the equilibrium lies to the left, favoring the undissociated form \([HX]\).

Dissociation of Acids

Dissociation is the process by which an acid breaks apart into ions when dissolved in water. For monoprotic acids like \(HX\), this involves the release of one proton \((H^+)\). The chemical equation representing the dissociation can be written as:
\[HX \rightleftharpoons H^+ + X^-\]Initially, we start with a concentration of the acid, while the concentrations of \(H^+\) and \(X^-\) are zero. As the acid dissociates, some of \(HX\) becomes \(H^+\) and \(X^-\), reaching an equilibrium state.

• The degree of dissociation is related to \(K_a\). Higher \(K_a\) means more dissociation.
• Understanding initial and equilibrium concentrations helps in calculating how much of each species is present at equilibrium.

pH Calculation

The pH is a scale used to specify the acidity or basicity of an aqueous solution, and is calculated using the formula:
\[\text{pH} = -\log[H^+]\]To find the pH of a solution, you need the concentration of \(H^+\) ions, which you can determine from the equilibrium concentrations.
In our example, after calculating \(x\), the equilibrium concentration of \(H^+\) is found to be 0.0036 M. Using this value:
\[\text{pH} = -\log(0.0036) \approx 2.44\]This tells us that our solution is acidic, as pH values less than 7 are considered acidic.