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Chapter 14: Problem 64

Calculate the percent dissociation for a \(0.22 M\) solution of chlorous acid \(\left(\mathrm{HClO}_{2}, K_{\mathrm{a}}=1.2 \times 10^{-2}\right)\)

Short Answer

Expert verified
The percent dissociation for a \(0.22 M\) solution of chlorous acid (HClO₂) is approximately 20.9%.

Step by step solution

01

Set up the equilibrium expression

Write the equilibrium expression for the dissociation of chlorous acid, HClO₂, which is a weak acid. The dissociation reaction of HClO₂ can be represented as: HClO₂ (aq) ⇌ H⁺ (aq) + ClO₂⁻ (aq) Set up an equilibrium expression using the equilibrium constant, Ka: Ka = \(\frac{[H^{+}][ClO_{2}^{-}]}{[HClO_{2}]}\)
02

Define the change in concentration

Let x represent the change in concentration of H⁺ and ClO₂⁻ ions. Since HClO₂ dissociates into H⁺ and ClO₂⁻, we know that for every mole of HClO₂ that dissociates, one mole of H⁺ and one mole of ClO₂⁻ are produced. Initial concentrations: [HClO₂] = 0.22 M [H⁺] = 0 [ClO₂⁻] = 0 Change in concentrations: [HClO₂] = -x [H⁺] = +x [ClO₂⁻] = +x Equilibrium concentrations: [HClO₂] = 0.22 - x [H⁺] = x [ClO₂⁻] = x
03

Substitute equilibrium concentrations into the Ka expression

Now substitute the equilibrium concentrations into the Ka expression: \(1.2\times10^{-2} = \frac{x^{2}}{(0.22-x)}\) Since Ka is relatively large, we can assume that x is small compared to 0.22. This allows us to simplify the equation as: \(1.2\times10^{-2} = \frac{x^{2}}{0.22}\)
04

Solve for x

To find the value of x, we need to solve the simplified equation: \(x^{2} = 1.2\times10^{-2}\times0.22\) Now, take the square root of both sides: \(x = \sqrt{1.2\times10^{-2}\times0.22}\) Calculating, we get: \(x \approx 0.046\) The value of x represents the concentration of H⁺ ions (and ClO₂⁻ ions) at equilibrium.
05

Calculate percent dissociation

Now that we have the concentration of H⁺ ions at equilibrium (x), we can calculate the percent dissociation using the following formula: Percent dissociation = \(\frac{[H^{+}\; at\; equilibrium]}{[initial\; concentration\; of\; HClO_{2}]}\times 100\) Percent dissociation = \(\frac{0.046}{0.22} \times 100\) Calculating, we get: Percent dissociation ≈ 20.9% Thus, the percent dissociation for a 0.22 M solution of chlorous acid (HClO₂) is approximately 20.9%.

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Most popular questions from this chapter

What are the major species present in \(0.250 M\) solutions of each of the following acids? Calculate the \(\mathrm{pH}\) of each of these solutions. a. \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) b. HCN

For propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\), determine the concentration of all species present, the \(\mathrm{pH}\), and the percent dissociation of a \(0.100 M\) solution.

Identify the Lewis acid and the Lewis base in each of the following reactions. a. \(\mathrm{B}(\mathrm{OH})_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{B}(\mathrm{OH})_{4}^{-}(a q)+\mathrm{H}^{+}(a q)\) b. \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q)\) c. \(\mathrm{BF}_{3}(g)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BF}_{4}^{-}(a q)\)

Given that the \(K_{\mathrm{a}}\) value for acetic acid is \(1.8 \times 10^{-5}\) and the \(K_{\mathrm{a}}\) value for hypochlorous acid is \(3.5 \times 10^{-8}\), which is the stronger base, \(\mathrm{OCl}^{-}\) or \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\) ?

Would you expect \(\mathrm{Fe}^{3+}\) or \(\mathrm{Fe}^{2+}\) to be the stronger Lewis acid? Explain.
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