What is the percent ionization of a 0.13 M formic acid solution in pure water, and how does it compare to the percent ionization in a solution containing 0.11 M potassium formate?

Verified step by step guidance

1. Identify the chemical equation for the ionization of formic acid: $\text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^-$. Formic acid is a weak acid, so it partially ionizes in solution.
2. Use the expression for the acid dissociation constant ($K_a$) for formic acid: $K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}$. Look up or be given the $K_a$ value for formic acid.
3. For the 0.13 M formic acid solution, set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of ions at equilibrium. Assume initial concentrations: $[\text{HCOOH}] = 0.13$ M, $[\text{H}^+] = 0$, $[\text{HCOO}^-] = 0$.
4. Calculate the percent ionization for the 0.13 M formic acid solution using the formula: $\text{Percent Ionization} = \left(\frac{[\text{H}^+]}{[\text{HCOOH}]_{\text{initial}}}\right) \times 100\%$. Use the equilibrium concentration of $[\text{H}^+]$ from the ICE table.
5. For the solution containing 0.11 M potassium formate, recognize that potassium formate is a salt that provides the common ion $\text{HCOO}^-$. Apply the common ion effect to the equilibrium, which will suppress the ionization of formic acid, and recalculate the percent ionization using a modified ICE table.