Problem 74 - Chemistry

Question

An acid HX is \(25 \%\) dissociated in water. If the equilibrium concentration of \(\mathrm{HX}\) is \(0.30 \mathrm{M}\), calculate the \(K_{\mathrm{a}}\) value for \(\mathrm{HX}\).

Step-by-Step Solution

Answer

The Ka value for HX is approximately \( 2.50 \times 10^{-2} \).

Step 1: Write the dissociation equation and Kа expression

First, write the equilibrium equation for the dissociation of HX and the expression for Ka: HX(aq) ↔ H+(aq) + X-(aq) Ka = [H+][X-] / [HX] Now we can use the given information to find the concentration of each species at equilibrium.

Step 2: Calculate the equilibrium concentrations of H+ and X-

We know that HX is 25% dissociated and the equilibrium concentration of HX is 0.30 M. We can use this information to calculate the amounts of HX that dissociated. Dissociated HX = 0.25 × 0.30 = 0.075 M As HX dissociates into H+ and X-, the concentrations of H+ and X- at equilibrium will be the same as the amount of HX that dissociated: [H+] = [X-] = 0.075 M

Step 3: Calculate the equilibrium concentration of HX

Since HX is 25% dissociated, 75% of it remains undissociated at equilibrium. To find the equilibrium concentration of HX, subtract the amount of HX that dissociated from the initial concentration: [HX]initial = 0.30 M [HX]dissociated = 0.075 M [HX]equilibrium = [HX]initial - [HX]dissociated = 0.30 - 0.075 = 0.225 M

Step 4: Calculate the Ka value

Now, we have the equilibrium concentrations of HX, H+, and X-. We can substitute these values into the Ka expression to find the Ka value: Ka = [H+][X-] / [HX] Ka = (0.075)(0.075) / (0.225) Ka = \( \frac{0.075^2}{0.225} \) Ka ≈ \( 2.50 \times 10^{-2} \) The Ka value for HX is approximately 2.50 x 10⁻².