Problem 75 - Chemistry An Atoms First Approach

Question

A \(0.15-M\) solution of a weak acid is \(3.0 \%\) dissociated. Calculate \(K_{\mathrm{a}}\)

Step-by-Step Solution

Answer

The equilibrium constant, \(K_a\), for the weak acid is approximately 1.39 × 10⁻⁴.

Step 1: Write the balanced equation for the dissociation of the weak acid

The general equation for the dissociation of a weak acid (HA) in water is: HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)

Step 2: Calculate the equilibrium concentrations of all species from the given data

First, we need to calculate the equilibrium concentrations of all species involved in the reaction. We are given the initial concentration of the weak acid (0.15 M) and its percentage of dissociation (3.0 %). Let x = the concentration of H3O+ and A- at equilibrium: x = 0.15 M × 3.0 % = 0.15 M × 0.03 = 0.0045 M So, the equilibrium concentrations are: [H3O+] = [A-] = 0.0045 M [HA] = 0.15 M - 0.0045 M = 0.1455 M

Step 3: Write the expression for the equilibrium constant, \(K_a\), and substitute equilibrium concentrations

The expression for \(K_a\) is: \(K_a\) = \(\frac{[H3O^+][A^-]}{[HA]}\) Now substitute the equilibrium concentrations we've found in Step 2: \(K_a\) = \(\frac{(0.0045)(0.0045)}{(0.1455)}\)

Step 4: Calculate the value of \(K_a\)

To find the value of \(K_a\), simply multiply the numerator, divide by the denominator, and solve: \(K_a\) = \(\frac{(0.0045)(0.0045)}{(0.1455)}\) = \(\frac{0.00002025}{0.1455}\) = 1.39 × 10⁻⁴ The equilibrium constant, \(K_a\), for the weak acid is approximately 1.39 × 10⁻⁴.