Problem 102 - Chemistry: An Atoms-Focused Approach

Question

Predict which solution in each pair below will have the lower pH. a. \(2.56 \times 10^{-2} \mathrm{M} \mathrm{HCl}\) or \(4.09 \times 10^{-2} \mathrm{MHBr}\) b. \(1.00 \times 10^{-5} M\) acctic acid \(\left(K_{\mathrm{a}}=1.76 \times 10^{-5}\right)\) or \(1.00 \times 10^{-5} M\) formic acid \(\left(K_{a}=1.77 \times 10^{-4}\right)\) c. \(22 \mathrm{mM} \mathrm{CH}_{3} \mathrm{NH}_{2}\left(\mathrm{p} K_{\mathrm{b}}=3.36\right)\) or \(22 \mathrm{mM}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\left(K_{\mathrm{b}}=5.9 \times 10^{-4}\right)\) d. \(158 \mathrm{mM} \mathrm{NH}_{3}\left(\mathrm{p} K_{\mathrm{b}}=4.75\right)\) or \(158 \mathrm{m} M\) acetic acid \(\left(p K_{a}=4.75\right)\) e. \(0.00395 M \mathrm{HNO}_{3}\) or \(0.00145 M \mathrm{HClO}_{4}\) f. \(2.05 \times 10^{-1} M\) propionic acid \(\left(K_{2}=1.4 \times 10^{-5}\right)\) or \(2.05 \times 10^{-1} M\) fluoroacetic acid \(\left(K_{2}=2.6 \times 10^{-3}\right)\) g. 375 m \(M\) pyridine \(\left(p K_{b}=8.77\right)\) or \(375 \mathrm{mM}\) aniline \(\left(\mathrm{p} K_{\mathrm{b}}=9.40\right)\) h. \(0.555 M \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\left(K_{2}=3 \times 10^{-3}\right)\) or \(0.355 M \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\left(K_{2}=1 \times 10^{-4}\right)\)

Step-by-Step Solution

Answer

a. \(HCl\) and \(HBr\) : The \(HBr\) solution will have a lower pH. b. Acetic acid and formic acid : The formic acid solution will have a lower pH. c. \(CH_3NH_2\) and \((CH_3)_2NH\) : The \(CH_3NH_2\) solution will have a lower pH. d. \(NH_3\) and acetic acid : The acetic acid solution will have a lower pH. e. \(HNO_3\) and \(HClO_4\) : The \(HNO_3\) solution will have a lower pH. f. Propionic acid and fluoroacetic acid : The fluoroacetic acid solution will have a lower pH. g. Pyridine and aniline : The aniline solution will have a lower pH. h. \(Fe(H_2O)_6^{3+}\) and \(Cr(H_2O)_6^{3+}\) : The \(Fe(H_2O)_6^{3+}\) solution will have a lower pH.

Step 1: a. Comparing \(\mathrm{HCl}\) and \(\mathrm{HBr}\) solutions

Both \(\mathrm{HCl}\) and \(\mathrm{HBr}\) are strong acids, meaning that they completely dissociate in water, so we can directly determine the concentration of \(\mathrm{H^+}\) from their initial concentrations. Since \(\mathrm{HBr}\) has a higher initial concentration (\(4.09 \times 10^{-2}\ \mathrm{M}\)) than \(\mathrm{HCl}\) (\(2.56 \times 10^{-2}\ \mathrm{M}\)), the \(\mathrm{HBr}\) solution will have a higher \(\mathrm{[H^+]}\) and thus a lower pH.

Step 2: b. Comparing acetic acid and formic acid solutions

Both acetic acid and formic acid are weak acids. To compare their pH values, we need to compare their \(K_a\) values. The higher the \(K_a\), the stronger the acid. Since formic acid has a higher \(K_a\) value (\(1.77 \times 10^{-4}\)) than acetic acid (\(1.76 \times 10^{-5}\)), the formic acid solution will have a higher \(\mathrm{[H^+]}\) and a lower pH.

Step 3: c. Comparing \(\mathrm{CH_3NH_2}\) and \(\mathrm{(CH_3)_2NH}\) solutions

Both \(\mathrm{CH_3NH_2}\) and \(\mathrm{(CH_3)_2NH}\) are weak bases. To compare their pH values, we need to compare their \(K_b\) values. The higher the \(K_b\), the stronger the base. However, remember we are comparing pH, not basicity. Since \(\mathrm{(CH_3)_2NH}\) has a higher \(K_b\) value (\(5.9 \times 10^{-4}\)) than \(\mathrm{CH_3NH_2}\) (\(pK_b = 3.36\) or \(K_b = 4.37 \times 10^{-4}\)), the \(\mathrm{(CH_3)_2NH}\) solution will have a higher \(\mathrm{[OH^-]}\) and a higher pH. Therefore, the \(\mathrm{CH_3NH_2}\) solution will have a lower pH.

Step 4: d. Comparing \(\mathrm{NH_3}\) and acetic acid solutions

\(\mathrm{NH_3}\) is a weak base and acetic acid is a weak acid. Comparing the given \(\mathrm{pK_b}\) value of \(\mathrm{NH_3}\) (\(4.75\)) to the given \(\mathrm{pK_a}\) value of acetic acid, we find that these two solutions have the same strength in their respective acid/base properties. However, since \(\mathrm{NH_3}\) is a base and acetic acid is an acid, the \(\mathrm{NH_3}\) solution will have a higher pH, whereas the acetic acid solution will have a lower pH.

Step 5: e. Comparing \(\mathrm{HNO_3}\) and \(\mathrm{HClO_4}\) solutions

Both \(\mathrm{HNO_3}\) and \(\mathrm{HClO_4}\) are strong acids that completely dissociate in water. Since \(\mathrm{HNO_3}\) has a higher initial concentration (\(0.00395\ \mathrm{M}\)) than \(\mathrm{HClO_4}\) (\(0.00145\ \mathrm{M}\)), the \(\mathrm{HNO_3}\) solution will have a higher \(\mathrm{[H^+]}\) and a lower pH.

Step 6: f. Comparing propionic acid and fluoroacetic acid solutions

Both propionic acid and fluoroacetic acid are weak acids. Since fluoroacetic acid has a higher \(K_a\) value (\(2.6 \times 10^{-3}\)) than propionic acid (\(1.4 \times 10^{-5}\)), the fluoroacetic acid solution will have a higher \(\mathrm{[H^+]}\) and a lower pH.

Step 7: g. Comparing pyridine and aniline solutions

Both pyridine and aniline are weak bases. Since aniline has a higher \(\mathrm{pK_b}\) value (\(9.40\)) than pyridine (\(8.77\)), it is a weaker base and thus has a lower \(\mathrm{[OH^-]}\). Therefore, the aniline solution will have a lower pH than the pyridine solution.

Step 8: h. Comparing \(\mathrm{Fe(H_2O)_6}^{3+}\) and \(\mathrm{Cr(H_2O)_6}^{3+}\) solutions

Both \(\mathrm{Fe(H_2O)_6}^{3+}\) and \(\mathrm{Cr(H_2O)_6}^{3+}\) are acidic species. Comparing their \(K_2\) values, we see that the \(\mathrm{Fe(H_2O)_6}^{3+}\) has a higher \(K_2\) value (\(3 \times 10^{-3}\)) than the \(\mathrm{Cr(H_2O)_6}^{3+}\) (\(1 \times 10^{-4}\)). Therefore, the \(\mathrm{Fe(H_2O)_6}^{3+}\) solution will have a higher \(\mathrm{[H^+]}\) and a lower pH.

Key Concepts

pH calculation

The pH is a measure of the acidity or basicity of a solution. It is calculated using the formula: \[ \text{pH} = -\log [\text{H}^+] \] Where \([\text{H}^+]\) represents the concentration of hydrogen ions in the solution. A lower pH value indicates a higher concentration of hydrogen ions, meaning the solution is more acidic. Conversely, a higher pH indicates a more basic solution. For strong acids like hydrochloric acid (HCl) and hydrobromic acid (HBr), which fully dissociate, the concentration of the acid directly gives the concentration of \(\text{H}^+\). For example, if you have a 0.01 M solution of HCl, the \([\text{H}^+]\) is 0.01 M, and the pH can be easily calculated as:\[ \text{pH} = -\log(0.01) = 2 \]This straightforward approach applies to strong acids and bases, but weak acids and bases require knowledge of their dissociation constants to find \([\text{H}^+]\) or \([\text{OH}^-]\).

Acid dissociation constant \(K_a\)

The acid dissociation constant, \(K_a\), is crucial in understanding the strength of a weak acid. It is a measure of how completely an acid dissociates in solution into its ions: \[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \] Here, \(K_a\) is defined as:\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \] A larger \(K_a\) value indicates a stronger acid, which means more \(\text{H}^+\) ions are produced, resulting in a lower pH. For instance, fluoroacetic acid, with a \(K_a\) of \(2.6 \times 10^{-3}\), is stronger than propionic acid, with a \(K_a\) of \(1.4 \times 10^{-5}\). This difference means fluoroacetic acid produces more hydrogen ions, leading to a more acidic solution with a lower pH. Weak acids like acetic and formic acids are common examples where \(K_a\) values help determine relative acidity.

Base dissociation constant \(K_b\)

Similar to \(K_a\), the base dissociation constant, \(K_b\), characterizes the strength of a weak base. It indicates the degree to which a base dissociates to form hydroxide ions (\(\text{OH}^-\)) in solution: \[ \text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^- \] \(K_b\) can be expressed as:\[ K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} \] A larger \(K_b\) value means a stronger base, corresponding to higher \([\text{OH}^-]\) and thus a higher pH. For instance, \((\text{CH}_3)_2\text{NH}\) has a \(K_b\) of \(5.9 \times 10^{-4}\), stronger than \(\text{CH}_3\text{NH}_2\), which makes the solution more basic and results in a higher pH. Understanding \(K_b\) is essential for evaluating the basicity of compounds like pyridine and aniline, as their relative strengths influence the pH of their solutions.