A 0.085 M solution of a monoprotic acid has a percent ionization of 0.59%. Determine the acid ionization constant (Ka) for the acid.

Verified step by step guidance

1. <insert step 1> Calculate the concentration of ionized acid using the percent ionization formula: \( \text{Percent Ionization} = \left( \frac{[\text{H}^+]}{[\text{HA}]} \right) \times 100 \). Here, [HA] is the initial concentration of the acid, and [H+] is the concentration of ionized acid.>
2. <insert step 2> Rearrange the formula to solve for [H+]: \( [\text{H}^+] = \frac{\text{Percent Ionization} \times [\text{HA}]}{100} \).>
3. <insert step 3> Substitute the given values into the equation: \( [\text{H}^+] = \frac{0.59 \times 0.085}{100} \).>
4. <insert step 4> Write the expression for the acid ionization constant \( K_a \) for a monoprotic acid: \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \). Since the acid is monoprotic, [H+] = [A-].>
5. <insert step 5> Substitute the values into the \( K_a \) expression: \( K_a = \frac{([\text{H}^+])^2}{[\text{HA}] - [\text{H}^+]} \). Use the calculated [H+] and initial [HA] to find \( K_a \).>