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Chapter 16: Problem 119

The equilibria \(\mathrm{OH}^{-}+\mathrm{HClO} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{ClO}^{-}\) and \(\mathrm{ClO}^{-}+\mathrm{HNO}_{2} \longrightarrow \mathrm{HClO}+\mathrm{NO}_{2}^{-}\) both lie to the right. Which of the following is a list of acids ranked in order of decreasing strength? (a) \(\mathrm{HClO}>\mathrm{HNO}_{2}>\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{ClO}^{-}>\mathrm{NO}_{2}^{-}>\mathrm{OH}^{-}\) (c) \(\mathrm{NO}_{2}^{-}>\mathrm{ClO}^{-}>\mathrm{OH}\) (d) \(\mathrm{HNO}_{2}>\mathrm{HClO}>\mathrm{H}_{2} \mathrm{O}\) (e) none of these

Short Answer

Expert verified
The correct answer is (d) \(\mathrm{HNO}_{2}>\mathrm{HClO}>\mathrm{H}_{2} \mathrm{O}\)

Step by step solution

01

Analyze the first equation

By looking at the first equilibrium \(\mathrm{OH}^{-}+\mathrm{HClO} \longrightarrow \mathrm{H}_{2}\mathrm{O}+\mathrm{ClO}^{-}\), since the reaction lies to the right, it means HClO is a stronger acid than H2O.
02

Analyze the second equation

By analyzing the second equilibrium \(\mathrm{ClO}^{-}+\mathrm{HNO}_{2}\longrightarrow \mathrm{HClO}+\mathrm{NO}_{2}^{-}\), once again, since the reaction lies to the right, it suggests HNO2 is a stronger acid than HClO.
03

Combine the results of both equations to form order

Combining the results from step 1 and step 2, the strongest acid is HNO2, then comes HClO, and finally H2O as the weakest.

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Most popular questions from this chapter

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

The pH of saturated \(\operatorname{Sr}(\text { OH })_{2}(\text { aq })\) is found to be 13.12 A \(10.0 \mathrm{mL}\) sample of saturated \(\operatorname{Sr}(\mathrm{OH})_{2}(\mathrm{aq})\) is diluted to \(250.0 \mathrm{mL}\) in a volumetric flask. A \(10.0 \mathrm{mL}\) sample of the diluted \(\operatorname{Sr}(\mathrm{OH})_{2}(\mathrm{aq})\) is transferred to a beaker, and some water is added. The resulting solution requires \(25.1 \mathrm{mL}\) of a \(\mathrm{HCl}\) solution for its titration. What is the molarity of this HCl solution?

Arrange the following 0.010 M solutions in order of increasing \(\mathrm{pH}: \mathrm{NH}_{3}(\mathrm{aq}), \mathrm{HNO}_{3}(\mathrm{aq}), \mathrm{NaNO}_{2}(\mathrm{aq})\) \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}), \quad \mathrm{NaOH}(\mathrm{aq}), \quad \mathrm{NH}_{4} \mathrm{CH}_{3} \mathrm{COO}(\mathrm{aq})\) \(\mathrm{NH}_{4} \mathrm{ClO}_{4}(\mathrm{aq})\)

For each of the following, identify the acids and bases involved in both the forward and reverse directions. (a) $\mathrm{HOBr}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{OBr}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}$ (b) $\mathrm{HSO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{SO}_{4}^{2-}+\mathrm{H}_{3} \mathrm{O}^{+}$ (c) $\mathrm{HS}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{S}+\mathrm{OH}^{-}$ (d) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O}$

\(\operatorname{In} 0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}),\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) is equal to \((\mathrm{a}) 0.050 \mathrm{M}\) (b) \(0.10 \mathrm{M} ;\) (c) \(0.11 \mathrm{M} ;\) (d) \(0.20 \mathrm{M}\).
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