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Chapter 16: Problem 29

What are \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right], \mathrm{pH},\) and \(\mathrm{pOH}\) of \(0.55 \mathrm{M}\) \(\mathrm{M} \mathrm{HClO}_{2} ?\)

Short Answer

Expert verified
The concentration of hydronium ions is 0.55 M, the concentration of hydroxide ions is \(1.82 \times 10^{-14} M\), the pH is approximately 0.26 and the pOH is approximately 13.74.

Step by step solution

01

Calculation of Hydronium Ion Concentration

Since HClO2 is a strong acid, it dissociates completely in water to form hydronium ions (H3O+) and chlorite ions (ClO2-). Therefore, the concentration of hydronium ions will be equal to the molarity of the HClO2, i.e. \( [\mathrm{H}_{3}\mathrm{O}^{+}]= 0.55 M\)
02

Calculation of Hydroxide Ion Concentration

The product of the concentrations of hydronium and hydroxide ions in water at 25°C is always \(1 \times 10^{-14}\). Therefore, use this relation to find the concentration of hydroxide ions: \(\left[\mathrm{OH}^{-}\right]=\frac{1 \times 10^{-14}}{\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]}\). Substitute values to get \(\left[\mathrm{OH}^{-}\right]\), which turns out to be \(1.82 \times 10^{-14} M\)
03

Calculation of pH and pOH

The pH is the negative logarithm (base 10) of the concentration of hydronium ions. Therefore, calculate it as: pH = -log(\( [\mathrm{H}_{3}\mathrm{O}^{+}] \)), which equals -log(0.55) or approximately 0.26. Similarly, the pOH is the negative logarithm (base 10) of the concentration of hydroxide ions: pOH = -log(\(\left[\mathrm{OH}^{-}\right]\)), which equals -log(\(1.82 \times 10^{-14}\)) or approximately 13.74

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Most popular questions from this chapter

The molecular solid \(\mathrm{I}_{2}(\mathrm{s})\) is only slightly soluble in water but will dissolve to a much greater extent in an aqueous solution of \(\mathrm{KI}\), because the \(\mathrm{I}_{3}^{-}\) anion forms. Write an equation for the formation of the \(I_{3}^{-}\) anion, and indicate the Lewis acid and Lewis base.

Explain the important distinctions between each pair of terms: (a) Bronsted- Lowry acid and base; (b) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\mathrm{pH} ;\) (c) \(K_{\mathrm{a}}\) for \(\mathrm{NH}_{4}^{+}\) and \(K_{\mathrm{b}}\) for \(\mathrm{NH}_{3} ;\) (d) leveling effect and electron- withdrawing effect.

Phosphorous acid is listed in Appendix D as a diprotic acid. Propose a Lewis structure for phosphorous acid that is consistent with this fact.

In \(0.10 \mathrm{M} \quad \mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{aq}), \quad\) (a) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.10 \mathrm{M}\) (b) \(\left[\mathrm{OH}^{-}\right]=0.10 \mathrm{M} ;(\mathrm{c}) \mathrm{pH}<7 ;(\mathrm{d}) \mathrm{pH}<13\).

\(\mathrm{CO}_{2}(\mathrm{g})\) can be removed from confined quarters (such as a spacecraft) by allowing it to react with an alkali metal hydroxide. Show that this is a Lewis acid-base reaction. For example, $$\mathrm{CO}_{2}(\mathrm{g})+\mathrm{LiOH}(\mathrm{s}) \longrightarrow \mathrm{LiHCO}_{3}(\mathrm{s})$$
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