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Chapter 13: Problem 115

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 \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} ?\)

Short Answer

Expert verified
The \(K_\mathrm{a}\) values for acetic acid and hypochlorous acid are \(1.8 \times 10^{-5}\) and \(3.5 \times 10^{-8}\), respectively. Since acetic acid is the stronger acid, its conjugate base, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\), is weaker than the conjugate base of hypochlorous acid, \(\mathrm{OCl}^{-}\). Therefore, the stronger base is \(\mathrm{OCl}^{-}\).

Step by step solution

01

Understand the relationship between \(K_\mathrm{a}\) and base strength

The \(K_\mathrm{a}\) values of conjugate acids are inversely proportional to the strength of their conjugate bases. Acids with a higher \(K_\mathrm{a}\) value will be stronger, which in turn will make their conjugate bases weaker.
02

Compare the \(K_\mathrm{a}\) values of acetic acid and hypochlorous acid

We are given the \(K_\mathrm{a}\) values for the following acids: - Acetic acid: \(1.8 \times 10^{-5}\) - Hypochlorous acid: \(3.5 \times 10^{-8}\) Since the higher \(K_\mathrm{a}\) value represents a stronger acid, we can determine that acetic acid is the stronger acid of the two, as \(1.8 \times 10^{-5} > 3.5 \times 10^{-8}\).
03

Determine the stronger conjugate base

Now that we know acetic acid is the stronger acid of the two, we can infer that its conjugate base, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\), is weaker than the conjugate base of hypochlorous acid, \(\mathrm{OCl}^{-}\). Therefore, the stronger base is \(\mathrm{OCl}^{-}\).

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Most popular questions from this chapter

A solution is prepared by adding 50.0 mL concentrated hydrochloric acid and \(20.0 \mathrm{mL}\) concentrated nitric acid to 300 mL water. More water is added until the final volume is 1.00 L. Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and the \(\mathrm{pH}\) for this solution. [Hint: Concentrated HCl is \(38 \%\) HCl (by mass) and has a density of 1.19 g/mL; concentrated HNO_ is 70.\% HNO_3 (by mass) and has a density of \(1.42 \mathrm{g} / \mathrm{mL} .]\)

How would you prepare \(1600 \mathrm{mL}\) of a \(\mathrm{pH}=1.50\) solution using concentrated (12 \(M\) ) HCl?

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction $$\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons{\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$$ is \(6.0 \times 10^{-3}\) a. Calculate the \(\mathrm{pH}\) of a \(0.10-M\) solution of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) b. Will a \(1.0-M\) solution of iron(II) nitrate have a higher or lower \(\mathrm{pH}\) than a \(1.0-M\) solution of iron(III) nitrate? Explain.

Making use of the assumptions we ordinarily make in calculating the pH of an aqueous solution of a weak acid, calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-6}-M\) solution of hypobromous acid (HBrO, \(K_{\mathrm{a}}=2 \times 10^{-9} \mathrm{J} .\) What is wrong with your answer? Why is it wrong? Without trying to solve the problem, explain what has to be included to solve the problem correctly.

Calculate the \(\mathrm{pH}\) of each of the following solutions of a strong acid in water. a. \(0.10 \mathrm{M} \mathrm{HCl}\) b. 5.0 M HCl c. \(1.0 \times 10^{-11} \mathrm{M} \mathrm{HCl}\)
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