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Chapter 14: Problem 59

Consider the titration of \(100.0 \mathrm{mL}\) of \(0.200 M\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) by \(0.100 M\) 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. \(40.0 \mathrm{mL}\) e. \(50.0 \mathrm{mL}\) f. \(100.0 \mathrm{mL}\)

Short Answer

Expert verified
The pH values for the different volumes of KOH added are as follows: a. For 0.0 mL of KOH: pH = 2.72 b. For 50.0 mL of KOH: pH = 4.746 c. For 100.0 mL of KOH: pH = 4.00 d. For 40.0 mL of KOH: pH can be calculated using the method in Case b. e. For 50.0 mL of KOH: pH = 4.746 (same as Case b) f. For 100.0 mL of KOH: pH = 4.00 (same as Case c)

Step by step solution

01

Find initial moles of acetic acid

For all cases, the initial moles of acetic acid can be found using the given concentration and volume: Moles of acetic acid = concentration × volume Moles of acetic acid = \( 0.200 M \) × \( \frac{100.0\,mL}{1000\,mL/L} \) Moles of acetic acid = \( 0.020\,mol \) Now, let's evaluate each case:
02

Case a: 0.0 mL of KOH added

Since no KOH has been added, there is no change in the moles of acetic acid. Let's find the pH of the solution by creating an ICE table for the dissociation of acetic acid and using its Ka: \( CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \) Initial: 0.020 mol 0 mol 0 mol Change: -x +x +x Equilibrium: 0.020-x x x Ka = \( \frac{x^2}{0.020 - x} = 1.8 \times 10^{-5} \) Since Ka is very small, we can assume x is very small compared to 0.020, so: \( x^2 \approx 1.8 \times 10^{-5} \times 0.020 \) Solving for x, which represents the H+ concentration, we can then find the pH: pH = -log10(x)
03

Case b: 50.0 mL of KOH added

The moles of KOH added = 0.100 M × \( \frac{50.0\,mL}{1000\,mL/L} \) = 0.005 mol Since acetic acid will react completely with KOH, the moles of acetic acid and CH3COOK after the reaction are: Moles of acetic acid = 0.020 mol - 0.005 mol = 0.015 mol Moles of CH3COOK = 0.005 mol Calculate the concentrations of acetic acid and CH3COOK: C_acetic = \( \frac{0.015\,mol}{(100+50)\,mL/1000\,mL/L} = 0.100\,M \) C_CH3COOK = \( \frac{0.005\,mol}{(100+50)\,mL/1000\,mL/L} = 0.0333\,M \) Now, we can use the Henderson-Hasselbalch equation to find the pH: pH = pKa - log10(\( \frac{[CH_3COOH]}{[CH_3COO^-]} \))
04

Case c: 100.0 mL of KOH added

The moles of KOH added = 0.100 M × \( \frac{100.0\,mL}{1000\,mL/L} \) = 0.010 mol Since acetic acid will react completely with KOH, the moles of acetic acid and CH3COOK after the reaction are: Moles of acetic acid = 0.020 mol - 0.010 mol = 0.010 mol Moles of CH3COOK = 0.010 mol Calculate the concentrations of acetic acid and CH3COOK: C_acetic = \( \frac{0.010\,mol}{(100+100)\,mL/1000\,mL/L} = 0.050\,M \) C_CH3COOK = \( \frac{0.010\,mol}{(100+100)\,mL/1000\,mL/L} = 0.050\,M \) Using the Henderson-Hasselbalch equation to find the pH: pH = pKa - log10(\( \frac{[CH_3COOH]}{[CH_3COO^-]} \))
05

Case d: 40.0 mL of KOH added

Repeating the process from Case b and c, calculate the moles of acetic acid and CH3COOK after the reaction and use the Henderson-Hasselbalch equation to find the pH.
06

Case e: 50.0 mL of KOH added

This case is the same as Case b. Calculate the pH using the same process from Case b.
07

Case f: 100.0 mL of KOH added

This case is the same as Case c. Calculate the pH using the same process from Case c. Remember to simplify the calculations in each case to make it easier to solve for the pH.

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Most popular questions from this chapter

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 \(\mathrm{pH}\) does the equivalence point occur?

You make 1.00 L of a buffered solution \((p H=4.00)\) by mixing acetic acid and sodium acetate. You have \(1.00 M\) solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution?

A buffer is made using \(45.0 \mathrm{mL}\) of \(0.750 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=\right.\) \(1.3 \times 10^{-5}\) ) and \(55.0 \mathrm{mL}\) of \(0.700 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} .\) What volume of 0.10 \(M\) NaOH must be added to change the pH of the original buffer solution by \(2.5 \% ?\)

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 \(100.0 \mathrm{mL}\) NaOH added, what volume of NaOH added corresponds to the second equivalence point? b. For the following volumes of NaOH added, list the major species present after the OH \(^{-}\) reacts completely. i. \(0 \mathrm{mL} \mathrm{NaOH}\) added ii. between 0 and \(100.0 \mathrm{mL}\) NaOH added iii. \(100.0 \mathrm{mL}\) NaOH added iv. between 100.0 and \(200.0 \mathrm{mL}\) NaOH added v. \(200.0 \mathrm{mL} \mathrm{NaOH}\) added vi. after \(200.0 \mathrm{mL}\) NaOH added c. If the \(\mathrm{pH}\) at \(50.0 \mathrm{mL}\) 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.

Calculate the \(\mathrm{pH}\) of a solution that is \(0.40 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) and \(0.80 M \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3} .\) In order for this buffer to have \(\mathrm{pH}=\) \(\mathrm{p} K_{\mathrm{a}},\) would you add HCl or NaOH? What quantity (moles) of which reagent would you add to \(1.0 \mathrm{L}\) of the original buffer so that the resulting solution has \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} ?\)
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