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Chapter 13: Problem 98

Calculate \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right],\) and the \(\mathrm{pH}\) of \(0.40 M\) solutions of each of the following amines (the \(K_{\mathrm{b}}\) values are found in Table 13-3). a. aniline b. methylamine

Short Answer

Expert verified
For aniline, \(\left[\mathrm{OH}^{-}\right] = x ≈ 1.3 \times 10^{-5} M\), \(\left[\mathrm{H}^{+}\right] ≈ 7.7 \times 10^{-10} M\), and \(\mathrm{pH} ≈ 9.11\). For methylamine, \(\left[\mathrm{OH}^{-}\right] = x ≈ 6.63 \times 10^{-3} M\), \(\left[\mathrm{H}^{+}\right] ≈ 1.5 \times 10^{-12} M\), and \(\mathrm{pH} ≈ 11.82\).

Step by step solution

01

Write the equilibrium expressions for both amines

We will write the equilibrium expressions for both amines as they react with water to form hydroxide ions and their respective conjugate acids. The general reaction is: Amine + H2O \(\rightleftharpoons\) Conjugate Acid + OH- For aniline (C6H5NH2): C6H5NH2 + H2O \(\rightleftharpoons\) C6H5NH3+ + OH- For methylamine (CH3NH2): CH3NH2 + H2O \(\rightleftharpoons\) CH3NH3+ + OH-
02

Set up the ICE table for the equilibrium expressions

We will set up an ICE (Initial, Change, Equilibrium) table for each amine to determine how the concentrations change during the reaction. For aniline: \[ \begin{array}{cccc} &{} & C6 H5 N H2 & & + & & H2 O & \rightleftharpoons & C6 H5 N H3^{+} & + & O H^{-} \\ I & & 0.40 & M & - & - & - & & 0 & M & & 0 & M \\ C & & - x & & - & - & - & & + x & & + x \\ E & & 0.40 - x & & - & - & - & & x & & x \end{array} \] For methylamine: \[ \begin{array}{cccc} &{} & C H3 N H2 & & + & & H2 O & \rightleftharpoons & C H3 N H3^{+} & + & O H^{-} \\ I & & 0.40 & M & - & - & - & & 0 & M & & 0 & M \\ C & & - x & & - & - & - & & + x & & + x \\ E & & 0.40 - x & & - & - & - & & x & & x \end{array} \]
03

Use the Kb values to solve for x, and then find [OH-]

For aniline, given \(K_b = 4.2 \times 10^{-10}\): \[K_b = \frac{[C6H5NH3^+][OH⁻]}{[C6H5NH2]} = \frac{x^2}{0.40-x}\] For methylamine, given \(K_b = 4.4 \times 10^{-4}\): \[K_b = \frac{[CH3NH3^+][OH⁻]}{[CH3NH2]} = \frac{x^2}{0.40-x}\] For each amine, we will solve for x, which represents the equilibrium concentration of OH⁻. Since \(K_b\) values are small, we can assume that \(x << 0.40\) and simplify the equations. For aniline: \[4.2 \times 10^{-10} \approx \frac{x^2}{0.40}\] Solve for x to find \(\left[OH^{-}\right]\). For methylamine: \[4.4 \times 10^{-4} \approx \frac{x^2}{0.40}\] Solve for x to find \(\left[OH^{-}\right]\).
04

Calculate [H+] and pH

Using the obtained OH⁻ concentration, we can find the H⁺ concentration using the relationship: \(K_w = \left[H^+\right]\left[OH^{-}\right]\), where \(K_w = 1.0 \times 10^{-14}\). Solve for the [H+] for both amines and use the formula to find the pH: \(pH = -\log\left([H^+]\right)\) Calculate the pH of both the aniline and methylamine solutions.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of a \(0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

The \(\mathrm{pH}\) of a \(0.016-M\) aqueous solution of \(p\) -toluidine \(\left(\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2}\right)\) is \(8.60 .\) Calculate \(K_{\mathrm{b}}\).

Calculate the \(\mathrm{pH}\) of each of the following solutions containing a strong acid in water. a. \(2.0 \times 10^{-2} M \mathrm{HNO}_{3}\) b. \(4.0 \mathrm{M} \mathrm{HNO}_{3}\) c. \(6.2 \times 10^{-12} \mathrm{M} \mathrm{HNO}_{3}\)

Using your results from Exercise \(133,\) place the species in each of the following groups in order of increasing base strength. a. \(\mathrm{OH}^{-}, \mathrm{SH}^{-}, \mathrm{SeH}^{-}\) b. \(\mathrm{NH}_{3}, \mathrm{PH}_{3}\) c. \(\mathrm{NH}_{3}, \mathrm{HONH}_{2}\)

Calculate the \(\mathrm{pH}\) of a \(0.200-M\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is $$\begin{aligned} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \\ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q) &+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\end{aligned}$$
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