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Chapter 21: Problem 63

Choose the strongest acid from each group. (a) \(\mathrm{HClO}, \mathrm{HBrO}, \mathrm{HIO}\) (b) \(\mathrm{HIO}, \mathrm{HIO}_{3}, \mathrm{HIO}_{4}\) (c) \(\mathrm{HIO}, \mathrm{HBrO}_{2}, \mathrm{HBrO}_{4}\)

Short Answer

Expert verified
Question: Identify the strongest acid from each group of given compounds: (a) \(\mathrm{HClO}, \mathrm{HBrO}, \mathrm{HIO}\) (b) \(\mathrm{HIO}, \mathrm{HIO}_{3}, \mathrm{HIO}_{4}\) (c) \(\mathrm{HIO}, \mathrm{HBrO}_{2}, \mathrm{HBrO}_{4}\) Answer: (a) \(\mathrm{HIO}\) (b) \(\mathrm{HIO}_{4}\) (c) \(\mathrm{HBrO}_{4}\)

Step by step solution

01

Identify the periodic trend in group (a) acids

In Group (a) \(\mathrm{HClO}, \mathrm{HBrO}, \mathrm{HIO}\), the central atoms are from the same group of the periodic table (Group 17 - Halogens) but belong to different periods. As we go down a group in the periodic table, the atomic size increases, which leads to a decrease in \(\%\) s-character in the central atom's hybridization. This means the central atom becomes less electronegative and has less ability to stabilize the negative charge that forms when the hydrogen ion (\(\mathrm{H}^{+}\)) is released. As a result, the acidity of these oxyacids increases down the group.
02

Choose the strongest acid from group (a)

Based on the periodic trends explained in Step 1, we can conclude that \(\mathrm{HIO}\) is the strongest acid in group (a) due to its larger central atom, Iodine, that is less electronegative compared to Chlorine and Bromine.
03

Compare the oxidation states in group (b) acids

In Group (b) \(\mathrm{HIO}, \mathrm{HIO}_{3}, \mathrm{HIO}_{4}\), all the acids have the same central atom, Iodine. In such cases, we can compare their acidity based on the oxidation state of the central atom. Higher oxidation states lead to increased acidity as it results in the central atom becoming more electronegative and better able to stabilize the negative charge formed upon the loss of H\(^+\). Here, the oxidation states of Iodine are +1 in \(\mathrm{HIO}\), +5 in \(\mathrm{HIO}_{3}\), and +7 in \(\mathrm{HIO}_{4}\).
04

Choose the strongest acid from group (b)

Based on the oxidation states explained in Step 3, we can conclude that \(\mathrm{HIO}_{4}\) is the strongest acid in group (b) due to its highest oxidation state of Iodine, which makes it the most electronegative and better able to stabilize the negative charge.
05

Compare the acids from different groups in group (c) acids

In Group (c) \(\mathrm{HIO}, \mathrm{HBrO}_{2}, \mathrm{HBrO}_{4}\), the central atoms belong to different groups in the periodic table. In such cases, we can use a combination of both the periodic trend of acidity and oxidation states of central atoms. The electronegativity decreases down a group, meaning Chlorine is more electronegative than Bromine, which in turn is more electronegative than Iodine. Thus, in general, oxyacids of Chlorine tend to be stronger acids than those of Bromine, and oxyacids of Bromine tend to be stronger acids than those of Iodine. However, a higher oxidation state of the central atom can increase the acidity even more in certain cases.
06

Choose the strongest acid from group (c)

In group (c), we have \(\mathrm{HIO}\) (Iodine with oxidation state +1), \(\mathrm{HBrO}_{2}\) (Bromine with oxidation state +3), and \(\mathrm{HBrO}_{4}\) (Bromine with oxidation state +7). Considering both the periodic trend of acidity and the oxidation state effects, we can conclude that \(\mathrm{HBrO}_{4}\) is the strongest acid in group (c) due to the highest oxidation state of Bromine, which makes it more electronegative than the other central atoms in group (c). In conclusion, the strongest acids from each group are: (a) \(\mathrm{HIO}\) (b) \(\mathrm{HIO}_{4}\) (c) \(\mathrm{HBrO}_{4}\)

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Most popular questions from this chapter

If an electrolytic cell producing fluorine uses a current of \(7.00 \times 10^{3} \mathrm{~A}\) (at \(10.0 \mathrm{~V})\), how many grams of fluorine gas can be produced in two days (assuming that the cell operates continuously at \(95 \%\) efficiency)?

Given $$\begin{array}{ll} \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) & K_{\mathrm{a}}=6.9 \times 10^{-1} \\ \mathrm{HF}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}_{2}-(a q) & K=2.7 \end{array}$$ calculate \(K\) for the reaction $$ 2 \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HF}_{2}^{-}(a q) $$

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When a solution of hydrogen bromide is prepared, \(1.283 \mathrm{~L}\) of \(\mathrm{HBr}\) gas at \(25^{\circ} \mathrm{C}\) and \(0.974\) atm is bubbled into \(250.0 \mathrm{~mL}\) of water. Assuming all the HBr dissolves with no volume change, what is the molarity of the hydrobromic acid solution produced?
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