Question

\(0.2\) molar solution of formic acid is \(3.2 \%\) ionized. Its ionization constant is (1) \(9.6 \times 10^{3}\) (2) \(2.1 \times 10^{4}\) (3) \(1.25 \times 10^{6}\) (4) \(4.8 \times 10^{5}\)

Short answer

The ionization constant is \(2.1 \times 10^{-4}\).

Step-by-step solution

Write the ionization reaction of formic acid

Formic acid (HCOOH) ionizes in water as follows:\[ \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- \]

Determine the concentration of ionized formic acid

Given the initial concentration of formic acid is 0.2 M and it is 3.2% ionized, we can find the concentration of ionized formic acid:\[ \text{Concentration of ionized HCOOH} = 0.2 \text{ M} \times \frac{3.2}{100} = 0.0064 \text{ M} \]

Calculate the equilibrium concentrations

At equilibrium, the concentrations of \( \text{H}^+ \) and \( \text{HCOO}^- \) will both be 0.0064 M. The concentration of undissociated formic acid will be:\[ \text{HCOOH}_{\text{eq}} = 0.2 \text{ M} - 0.0064 \text{ M} = 0.1936 \text{ M} \]

Apply the ionization constant expression

The ionization constant (\( K_{a} \)) expression for formic acid is:\[ K_{a} = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \]Substitute the equilibrium concentrations into this expression:\[ K_{a} = \frac{(0.0064 \text{ M})(0.0064 \text{ M})}{0.1936 \text{ M}} \]

Calculate the ionization constant

Perform the calculation to find the ionization constant:\[ K_{a} = \frac{0.00004096}{0.1936} = 2.12 \times 10^{-4} \]

Identify the correct answer

Compare the calculated ionization constant with the options given in the problem. The closest value is option (2) \(2.1 \times 10^{-4}\).

Key concepts

acid-base equilibrium

Understanding acid-base equilibrium is crucial for solving ionization problems involving acids like formic acid. An acid-base equilibrium involves the ionization of an acid (in this case, formic acid) in water, where the acid donates a proton (H+) to the water, resulting in the formation of its conjugate base and hydronium ions (H3O+). In the given problem, formic acid (HCOOH) ionizes as follows: \[ \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- \] This equation shows the reversible, dynamic process where formic acid dissociates into hydrogen ions and formate ions (HCOO^-). At equilibrium, the rates of the forward and reverse reactions are equal, and the concentration of each species remains constant over time. This equilibrium position is quantified by the ionization constant (Ka), which is a key factor in understanding the strength and behavior of acids in solution.

ionization constant calculation

To calculate the ionization constant ( \(K_a\) ), we use the equilibrium concentrations of the ions produced and the remaining acid. The ionization constant expression for formic acid is given by: \[ K_{a} = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]} \] Given in the problem, the formic acid solution is 0.2 M, and 3.2% ionized. This means: \[0.2 M \times \frac{3.2}{100} = 0.0064 M\] of formic acid has ionized into \([\text{H}^+]\) and \([\text{HCOO}^-] \). The concentration of undissociated formic acid at equilibrium is \[0.2 M - 0.0064 M = 0.1936 M\]. Using these equilibrium concentrations in the \(K_a\) expression: \[ K_{a} = \frac{(0.0064 \text{ M})(0.0064 \text{ M})}{0.1936 \text{ M}} = 2.12 \times 10^{-4}\] This value matches closely with option (2) in the exercise, confirming our calculations and understanding of the ionization constant.

molarity

Molarity (M) is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per liter of solution. In this problem, the initial concentration of formic acid is given as 0.2 M, indicating there are 0.2 moles of HCOOH in every liter of solution. To calculate changes in concentrations due to ionization, we used molarity. For ionized formic acid: \[0.2 M \times \frac{3.2}{100} = 0.0064 M\] This means that 0.0064 mol of formic acid has ionized per liter of solution, leading to 0.0064 M each of \([\text{H}^+]\) and \([\text{HCOO}^-] \). Understanding molarity helps in further calculations of equilibrium concentrations and ensures the correct application of the ionization constant formula. Using molarity as a foundation, we can easily determine the extent of ionization and compute the equilibrium conditions accurately, making it a fundamental concept in solving such chemical problems.