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Chapter 13: Problem 103

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 \(K_{\mathrm{b}}\).

Short Answer

Expert verified
The base dissociation constant (Kb) of the p-toluidine solution is approximately \(9.96 \times 10^{-10}\).

Step by step solution

01

Calculate the hydroxide ion concentration

Since we are given the pH value of the solution (8.60), we can calculate the corresponding pOH value as: pOH = 14 - pH pOH = 14 - 8.60 = 5.4 Now we can calculate the hydroxide ion concentration (OH⁻) using the pOH value: \[\text{OH}^- = 10^{-\text{pOH}}\] \[ [\text{OH}^-] = 10^{-5.4}\] From this calculation, we find that the hydroxide ion concentration is approximately: \[ [\text{OH}^-] \approx 3.98 \times 10^{-6} \text{ M} \]
02

Write the dissociation reaction for p-toluidine

The dissociation reaction for p-toluidine can be represented as: \[\text{CH}_{3}\text{C}_{6}\text{H}_{4}\text{NH}_{2} + \text{H}_{2}\text{O} \rightleftarrows \text{CH}_{3}\text{C}_{6}\text{H}_{4}\text{NH}_{3}^{+} + \text{OH}^-\]
03

Calculate the change in concentration of reactants and products during dissociation

Let x be the concentration of dissociated p-toluidine, which is also equal to the concentration of hydroxide ions (OH-) at equilibrium. As a result, we have: [CH3C6H4NH3+ ] = [OH-] = x Since the initial concentration of p-toluidine (0.016 M) is much larger than the equilibrium concentration of hydroxide ions (about 3.98 × 10⁻⁶ M), we can assume that the change in p-toluidine concentration is negligible, and the equilibrium concentration is still approximately equal to 0.016 M.
04

Set up the Kb expression and find its value

The Kb expression for the dissociation reaction of p-toluidine is given by: \[\text{Kb} = \frac{[\text{CH}_{3}\text{C}_{6}\text{H}_{4}\text{NH}_{3}^{+}] \times [\text{OH}^-]}{[\text{CH}_{3}\text{C}_{6}\text{H}_{4}\text{NH}_{2}]}\] Substituting the values obtained in the previous steps, we have: \[\text{Kb} = \frac{(3.98 \times 10^{-6})^{2}}{0.016}\] Now, we can calculate the Kb value: \[\text{Kb} \approx 9.96 \times 10^{-10}\] So, the base dissociation constant (Kb) of the p-toluidine solution is approximately 9.96 × 10⁻¹⁰.

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Most popular questions from this chapter

A \(2.14-\mathrm{g}\) sample of sodium hypoiodite is dissolved in water to make 1.25 L of solution. The solution \(\mathrm{pH}\) is \(11.32 .\) What is \(K_{\mathrm{b}}\) for the hypoiodite ion?

An acid HX is \(25 \%\) dissociated in water. If the equilibrium concentration of HX is 0.30 \(M,\) calculate the \(K_{\mathrm{a}}\) value for HX.

Consider a \(0.67-M\) solution of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) a. Which of the following are major species in the solution? i. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) iii. OH \(^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) b. Calculate the \(p\) H of this solution.

A \(0.20-M\) sodium chlorobenzoate \(\left(\mathrm{NaC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution has a pH of \(8.65 .\) Calculate the pH of a 0.20- \(M\) chlorobenzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution.

For solutions of the same concentration, as acid strength increases, indicate what happens to each of the following (increases, decreases, or doesn't change). a. \(\left[\mathrm{H}^{+}\right]\) b. \(\mathrm{pH}\) c. \(\left[\mathrm{OH}^{-}\right]\) d. pOH \(\mathbf{e} . K_{\mathrm{a}}\)
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