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Chapter 16: Problem 58

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

Expert verified
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

01

Write the dissociation equation for chlorous acid.

The dissociation equation for chlorous acid is as follows: \[ \mathrm{HClO_2} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{ClO_2^-} \]
02

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

The expression for Ka is given as follows: \[ K_a = \frac{[\mathrm{H_3O^+}] [\mathrm{ClO_2^-}]}{[\mathrm{HClO_2}]} \]
03

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
04

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: \[ 1.1 \times 10^{-2} = \frac{x^2}{0.0125 - x} \]
05

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: \[ x = \sqrt{1.1 \times 10^{-2} (0.0125)} \approx 0.001235 \]
06

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⁻).

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