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Chapter 14: Problem 66

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
\[x \approx 0.04\text{ M}\] #tag_title# (Step 5: Calculate the percent dissociation)#tag_content# Percent dissociation = \(\frac{[H^+]}{[HA]_{initial}}\) x 100 Percent dissociation ≈ \(\frac{0.04}{0.22}\times 100 \approx 18.18 \%\)

Step by step solution

01

(Step 1: Write the Ka expression for HClO₂)

The dissociation reaction for HClO₂ is given by: HClO₂ ↔ H⁺ + ClO₂⁻ The Ka expression is: Ka = \(\frac{[H^+][ClO_2^-]}{[HClO_2]}\)
02

(Step 2: Set up the ICE table)

Set up the Initial, Change, and Equilibrium (ICE) table: \(|\quad \quad| [HClO_2] \quad | [H^+] \quad | [ClO_2^-] \quad\) \(---|---|---\) \(Initial| 0.22\ | 0\ \quad \quad | 0\quad \quad \ \) \(---|---|---\) \(Change\ | -x\ \quad\ | +x\ \quad\ | +x\quad \ \) \(---|---|---\) \(Equilibrium | 0.22-x\ | x\ \quad \ | x\quad \ \)
03

(Step 3: Substitute the equilibrium concentrations into the Ka expression)

Substitute equilibrium concentrations of H⁺, ClO₂⁻, and HClO₂, from the ICE table, into the Ka expression: \(1.2 \times 10^{-2} = \frac{x \cdot x}{0.22 - x}\)
04

(Step 4: Solve for x - the concentration of H⁺ ions)

In most cases, when the Ka value is very small, x can be considered negligible compared to the initial concentration (0.22 in this case). By making this approximation, we get: \(1.2 \times 10^{-2} \approx \frac{x^2}{0.22}\) Now, solve for x: \[x = \sqrt{0.22\times1.2 \times 10^{-2}}\]

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Most popular questions from this chapter

What are the major species present in the following mixtures of bases? a. \(0.050 \mathrm{M} \mathrm{NaOH}\) and \(0.050 \mathrm{M} \mathrm{LiOH}\) b. \(0.0010 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) and \(0.020 \mathrm{M} \mathrm{RbOH}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

Isocyanic acid (HNCO) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation \(2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\).

Consider a \(0.67-M\) solution of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\). a. Which of the following are major species in the solution? i. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) iii. \(\mathrm{OH}^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\) b. Calculate the \(\mathrm{pH}\) of this solution.

The following illustration displays the relative number of species when an acid, HA, is added to water. a. Is HA a weak or strong acid? How can you tell? b. Using the relative numbers given in the illustration, determine the value for \(K_{\mathrm{a}}\) and the percent dissociation of the acid. Assume the initial acid concentration is \(0.20 \mathrm{M}\).

Write out the stepwise \(K_{\mathrm{a}}\) reactions for citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\), a triprotic acid.
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