Question

The acid-dissociation constant for chlorous acid \(\left(\mathrm{HClO}_{2}\right)\) is \(1.1 \times 10^{-2} .\) Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{ClO}_{2}^{-}\) and \(\mathrm{HClO}_{2}\) at equilibrium if the initial concentration of \(\mathrm{HClO}_{2}\) is 0.0125 \(\mathrm{M} .\)

Short answer

The equilibrium concentrations for the dissociation of chlorous acid are approximately 0.011265 M (HClO₂), 0.001235 M (H₃O⁺), and 0.001235 M (ClO₂⁻).

Step-by-step solution

Write the dissociation equation for chlorous acid.

The dissociation equation for chlorous acid is as follows: \[ \mathrm{HClO_2} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{ClO_2^-} \]

Write the expression for the acid-dissociation constant, Ka.

The expression for Ka is given as follows: \[ K_a = \frac{[\mathrm{H_3O^+}] [\mathrm{ClO_2^-}]}{[\mathrm{HClO_2}]} \]

Define the initial and equilibrium concentrations.

Let's denote the initial concentration of HClO₂ as C (in our case 0.0125M). Initially, we have no H₃O⁺ and ClO₂⁻ ions. As the reaction proceeds, x moles of HClO₂ will dissociate, and as per the stoichiometry, x moles each of H₃O⁺ and ClO₂⁻ will be formed. So, the equilibrium concentrations will be as follows: [ HClO₂ ] = C - x [ H₃O⁺ ] = x [ ClO₂⁻ ] = x

Substitute equilibrium concentrations in the Ka expression

Plugging in the expressions for the equilibrium concentrations into the Ka expression, we get: \[ K_a = \frac{x^2}{C-x} \] Now, we know that Ka for chlorous acid is 1.1*10⁻². Therefore: \[ 1.1 \times 10^{-2} = \frac{x^2}{0.0125 - x} \]

Solve the quadratic equation for x

Now, we need to solve the quadratic equation for x, which might be simplified by multiplying both sides by (0.0125 - x) to get rid of the denominator: \[ x^2 = 1.1 \times 10^{-2} (0.0125 - x) \] This quadratic equation can be challenging to solve analytically, so using either a numerical solver or an appropriate approximation is recommended. In this case, since Ka is low, x is going to be substantially smaller than the initial concentration (C). Therefore, we can approximate (0.0125 - x) ≈ 0.0125. Now the equation can be written as: \[ x^2 = 1.1 \times 10^{-2} (0.0125) \] Solving for x: \[ x = \sqrt{1.1 \times 10^{-2} (0.0125)} \approx 0.001235 \]

Calculate the equilibrium concentrations of all species.

Now that we have the x value, we can calculate the equilibrium concentrations: [ HClO₂ ] = C - x ≈ 0.0125 - 0.001235 ≈ 0.011265M [ H₃O⁺ ] = x ≈ 0.001235 M [ ClO₂⁻ ] = x ≈ 0.001235 M So, the equilibrium concentrations are approximately 0.011265 M (HClO₂), 0.001235 M (H₃O⁺), and 0.001235 M (ClO₂⁻).

Key concepts

Chlorous Acid

Chlorous acid, denoted by the chemical formula \( \mathrm{HClO_2} \), is a weak acid. It is part of the family of oxoacids where chlorine is bonded to oxygen. The weakness of chlorous acid is due to its partial ionization in aqueous solutions.
When dissolved in water, chlorous acid partially dissociates to form hydronium ions \( \mathrm{H_3O^+} \) and chlorite ions \( \mathrm{ClO_2^-} \). This behavior is characteristic of weak acids, which do not completely dissolve into ions in water. Instead, an equilibrium exists between the undissociated molecules and the ions formed.
The degree of ionization is quantified by the acid-dissociation constant \( K_a \), which in the case of chlorous acid is given as \( 1.1 \times 10^{-2} \). This value indicates that chlorous acid is relatively stronger compared to many other weak acids, even though it is still not strong enough to fully dissociate like strong acids do. Understanding this partial dissociation is key to calculating equilibrium concentrations in solutions.

Equilibrium Concentrations

In the realm of chemistry, equilibrium concentrations represent the steady-state concentrations of reactants and products in a reaction that has reached equilibrium. For chlorous acid, equilibrium is attained when the rate of dissociation of \( \mathrm{HClO_2} \) into \( \mathrm{H_3O^+} \) and \( \mathrm{ClO_2^-} \) ions equals the rate of recombination of these ions back into \( \mathrm{HClO_2} \).
Calculating equilibrium concentrations involves using the initial concentrations and the changes that occur as the system reaches equilibrium. In this particular case, the initial concentration of \( \mathrm{HClO_2} \) was 0.0125 M and initially no \( \mathrm{H_3O^+} \) or \( \mathrm{ClO_2^-} \) ions are present. Let \( x \) represent the molar concentration of \( \mathrm{H_3O^+} \) and \( \mathrm{ClO_2^-} \) at equilibrium, reflecting the amount dissociated.
Thus, the equilibrium concentration of \( \mathrm{HClO_2} \) is \( 0.0125 - x \), and both \( \mathrm{H_3O^+} \) and \( \mathrm{ClO_2^-} \) are \( x \). These are crucial to substitute into the expression of \( K_a \) to find numerical solutions.

Quadratic Equation

Solving equilibrium problems often requires dealing with quadratic equations, especially in reactions involving weak acids like chlorous acid. When calculating equilibrium concentrations using the acid-dissociation constant equation \( K_a = \frac{x^2}{C-x} \), where \( C \) is the initial concentration of the acid, a quadratic equation emerges.
Formally, if we let \( x \) be the change in concentration due to the dissociation process, the equation to solve is: \( x^2 = K_a(C-x) \). Simplifying and finding \( x \) involves solving this quadratic equation by rearranging it to the standard form \( ax^2 + bx + c = 0 \). However, a practical approach is often taken in such cases.
Given that \( K_a \) is relatively small compared to \( C \), it is usually safe to approximate \( (C-x) \) to \( C \), substantially simplifying the math to \( x^2 = K_a \, C \). Consequently, \( x \) can be solved by the simple square root: \( x = \sqrt{K_a \, C} \).
This approximation allows a quick solution to find \( x \), after which equilibrium concentrations can be easily computed.