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Chapter 15: Problem 59

Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) by \(0.100 M \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of KOH have been added. a. \(0.0 \mathrm{~mL}\) b. \(50.0 \mathrm{~mL}\) c. \(100.0 \mathrm{~mL}\) d. \(150.0 \mathrm{~mL}\) e. \(200.0 \mathrm{~mL}\) f. \(250.0 \mathrm{~mL}\)

Short Answer

Expert verified
The pH values after adding the specified amounts of KOH are: a. 2.90 b. 4.74 c. 4.74 d. 4.74 e. 12.74 f. 13.62

Step by step solution

01

- Calculate Initial Moles of Acetic Acid

First, we will calculate the initial moles of acetic acid (CH3COOH) present in the solution. We are given the initial volume (100.0 mL) and concentration (0.200 M). Moles of acetic acid = volume × concentration Moles of acetic acid = 100.0 mL × 0.200 mol/L = 20.0 mmol
02

- Calculate Moles of KOH Added at Each Step

Next, we calculate the moles of KOH added at each step, using the given volumes and their 0.100 M concentration. a. 0.0 mL: 0.0 mmol KOH b. 50.0 mL: 5.0 mmol KOH c. 100.0 mL: 10.0 mmol KOH d. 150.0 mL: 15.0 mmol KOH e. 200.0 mL: 20.0 mmol KOH f. 250.0 mL: 25.0 mmol KOH
03

- Determine the Reaction between Acetic Acid and KOH

The reaction between acetic acid and KOH can be written as follows: CH3COOH + KOH -> CH3COO- + H2O Notice that the reaction produces acetate ions (CH3COO-) which are the conjugate base of acetic acid.
04

- Calculate the Amounts of Remaining Acid and Produced Conjugate Base

We will now calculate the amounts of remaining acetic acid and produced acetate ion at each step. a. 0.0 mL: 20.0 mmol CH3COOH, 0.0 mmol CH3COO- b. 50.0 mL: 15.0 mmol CH3COOH, 5.0 mmol CH3COO- c. 100.0 mL: 10.0 mmol CH3COOH, 10.0 mmol CH3COO- d. 150.0 mL: 5.0 mmol CH3COOH, 15.0 mmol CH3COO- e. 200.0 mL: 0.0 mmol CH3COOH, 20.0 mmol CH3COO- f. 250.0 mL: 0.0 mmol CH3COOH, 25.0 mmol CH3COO-
05

- Calculate Concentrations of the Species in the Solution

Now, we need to calculate the concentrations of the acetic acid, acetate ions, and added OH- ions in the solution for each step, dividing the moles by the total volume of the solution. For the acetic acid and acetate ions, the total volume can be found as the sum of the volume of the acetic acid solution and the volume of the added KOH. For OH- ion, the volume is the volume of added KOH.
06

- Calculate the pH Using the Acid Equilibrium Expression

Finally, we can calculate the pH using the equilibrium expression for dissociation of acetic acid: \(K_a = \frac{[\mathrm{CH3COO^-}][\mathrm{H^+}]}{[\mathrm{CH3COOH}]} = 1.8 \times 10^{-5}\) For each step, solve for [H+] and then calculate the pH as pH = -log[H+]. a. pH ≈ 2.90 b. pH ≈ 4.74 c. pH ≈ 4.74 d. pH ≈ 4.74 e. pH ≈ 12.74 f. pH ≈ 13.62 So, the pH values after adding the specified amounts of KOH are: a. 2.90 b. 4.74 c. 4.74 d. 4.74 e. 12.74 f. 13.62

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Most popular questions from this chapter

In the titration of \(50.0 \mathrm{~mL}\) of \(1.0 \mathrm{M}\) methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) \(\left(K_{\mathrm{b}}=4.4 \times 10^{-4}\right)\), with \(0.50 \mathrm{M} \mathrm{HCl}\), calculate the \(\mathrm{pH}\) under the following conditions. a. after \(50.0 \mathrm{~mL}\) of \(0.50 M \mathrm{HCl}\) has been added b. at the stoichiometric point

When a diprotic acid, \(\mathrm{H}_{2} \mathrm{~A}\), is titrated with \(\mathrm{NaOH}\), the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added, what volume of \(\mathrm{NaOH}\) added corresponds to the second equivalence point? b. For the following volumes of \(\mathrm{NaOH}\) added, list the major species present after the \(\mathrm{OH}^{-}\) reacts completely. i. \(0 \mathrm{~mL}\) NaOH added ii. between 0 and \(100.0 \mathrm{~mL}\) NaOH added iii. \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added iv. between \(100.0\) and \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added v. \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added vi. after \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added c. If the \(\mathrm{pH}\) at \(50.0 \mathrm{~mL} \mathrm{NaOH}\) added is \(4.0\) and the \(\mathrm{pH}\) at \(150.0 \mathrm{~mL} \mathrm{NaOH}\) added is \(8.0\), determine the values \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for the diprotic acid.

Consider a buffered solution containing \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(\mathrm{CH}_{3} \mathrm{NH}_{2}\). Which of the following statements concerning this solution is(are) true? \(\left(K_{\mathrm{a}}\right.\) for \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}=2.3 \times 10^{-11}\).) a. A solution consisting of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(0.10 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) would have a higher buffering capacity than one containing \(1.0 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(1.0 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\). b. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then the \(\mathrm{pH}\) is larger than the \(\mathrm{p} K_{\mathrm{a}}\) value. c. Adding more \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\right]\) to the initial buffer solution will decrease the \(\mathrm{pH}\). d. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]<\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then \(\mathrm{pH}<3.36\). e. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then \(\mathrm{pH}=10.64\).

What concentration of \(\mathrm{NH}_{4} \mathrm{Cl}\) is necessary to buffer a \(0.52-M\) \(\mathrm{NH}_{3}\) solution at \(\mathrm{pH}=9.00 ?\left(K_{\mathrm{b}}\right.\) for \(\left.\mathrm{NH}_{3}=1.8 \times 10^{-5} .\right)\)

A \(0.400-M\) solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?
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