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Chapter 13: Problem 114

Arrange the following 0.10 \(M\) solutions in order from most acidic to most basic. See Appendix 5 for \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values. $$\mathrm{CaBr}_{2}, \quad \mathrm{KNO}_{2}, \quad \mathrm{HClO}_{4}, \quad \mathrm{HNO}_{2}, \quad \mathrm{HONH}_{3} \mathrm{ClO}_{4}$$

Short Answer

Expert verified
The order of the given solutions from most acidic to most basic is: \(\mathrm{HClO}_{4}\), \(\mathrm{HNO}_{2}\), \(\mathrm{CaBr}_{2}\), \(\mathrm{HONH}_{3}\mathrm{ClO}_{4}\), and \(\mathrm{KNO}_{2}\).

Step by step solution

01

Identify the Type of Solution

First, we need to identify whether each given solution is acidic or basic. 1. \(\mathrm{CaBr}_{2}\) - It is a salt of weak base (\(\mathrm{Br^-}\)) and strong acid (\(\mathrm{Ca^{2+}}\)), so it will be acidic. 2. \(\mathrm{KNO}_{2}\) - It is a salt of weak acid (\(\mathrm{HNO}_{2}\)) and strong base (\(\mathrm{K^+}\)), so it will be basic. 3. \(\mathrm{HClO}_{4}\) - It is a strong acid, so it will be acidic. 4. \(\mathrm{HNO}_{2}\) - It is a weak acid, so it will be acidic. 5. \(\mathrm{HONH}_{3} \mathrm{ClO}_{4}\) - It is a salt of strong acid (\(\mathrm{HClO}_{4}\)) and weak base (\(\mathrm{HONH}_{3}\)), so it will be acidic. Now we know that out of these 5 given solutions, 4 are acidic and 1 is basic.
02

Compare Ka Values for Acidic Solutions

Now, let's compare \(K_{\mathrm{a}}\) values for the acidic solutions using Appendix 5. 1. \(\mathrm{CaBr}_{2}\) - \(K_{\mathrm{a}}\) not required directly, as Br is a weak base, and it will produce an acidic solution. 2. \(\mathrm{HClO}_{4}\) - \(K_{\mathrm{a}} > 1\) 3. \(\mathrm{HNO}_{2}\) - \(K_{\mathrm{a}} = 4.5 \times 10^{-4}\) 4. \(\mathrm{HONH}_{3} \mathrm{ClO}_{4}\) - \(K_{\mathrm{a}}\) not required directly, as HONH3 is a weak base, and it will produce an acidic solution. Out of these acidic solutions, \(\mathrm{HClO}_{4}\) has the highest \(K_{\mathrm{a}}\) value, followed by \(\mathrm{HNO}_{2}\).
03

Compare Kb Values for Basic Solutions

Now, let's compare \(K_{\mathrm{b}}\) values for the basic solution using Appendix 5. 1. \(\mathrm{KNO}_{2}\) - \(K_{\mathrm{b}} = 2.4 \times 10^{-2}\) Since we have only one basic solution, we don't need to compare it with others; we just need to place it properly in the final order.
04

Arrange the Solutions in the Order of Acidity

Based on the above analysis, we can arrange the given solutions in the following order, from most acidic to most basic: 1. \(\mathrm{HClO}_{4}\) (strong acid) 2. \(\mathrm{HNO}_{2}\) (weak acid) 3. \(\mathrm{CaBr}_{2}\) (acidic due to \(\mathrm{Br^-}\), a weak base) 4. \(\mathrm{HONH}_{3}\mathrm{ClO}_{4}\) (acidic due to \(\mathrm{HONH}_{3}\), a weak base) 5. \(\mathrm{KNO}_{2}\) (basic, due to \(\mathrm{NO}_{2^-}\), a weak acid)

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Most popular questions from this chapter

Identify the Lewis acid and the Lewis base in each of the following reactions. a. \(\mathrm{B}(\mathrm{OH})_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{B}(\mathrm{OH})_{4}^{-}(a q)+\mathrm{H}^{+}(a q)\) b. \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q)\) c. \(\mathrm{BF}_{3}(g)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BF}_{4}^{-}(a q)\)

A certain acid, HA, has a vapor density of \(5.11 \mathrm{g} / \mathrm{L}\) when in the gas phase at a temperature of \(25^{\circ} \mathrm{C}\) and a pressure of 1.00 atm. When 1.50 g of this acid is dissolved in enough water to make 100.0 mL of solution, the \(p\) H is found to be \(1.80 .\) Calculate \(K_{\mathrm{a}}\) for the acid.

For propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right),\) determine the concentration of all species present, the \(\mathrm{pH}\), and the percent dissociation of a 0.100-M solution.

A solution is made by adding \(50.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) to \(50.0 \mathrm{mL}\) of \(1.00 \times 10^{-3} \mathrm{M} \mathrm{HCl}\) a. Calculate the \(p\) H of the solution. b. Calculate the acetate ion concentration.

The pH of \(1.0 \times 10^{-8} M\) hydrochloric acid is not \(8.00 .\) The correct \(\mathrm{pH}\) can be calculated by considering the relationship between the molarities of the three principal ions in the solution \(\left(\mathrm{H}^{+}, \mathrm{Cl}^{-}, \text {and } \mathrm{OH}^{-}\right) .\) These molarities can be calculated from algebraic equations that can be derived from the considerations given below. a. The solution is electrically neutral. b. The hydrochloric acid can be assumed to be \(100 \%\) ionized. c. The product of the molarities of the hydronium ions and the hydroxide ions must equal \(K_{\mathrm{w}}\) Calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-8}-M\) HCl solution.
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