Buffers: Calculate pH of a Buffer Solution, and Calculate the pH of a Buffer Solution After Adding Strong Acid
Question
A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 × 10−5 as Ka for acetic acid, and solve using an ICE table approach.
- What is the pH of the solution?
- What is the pH of a solution that results when 0.500 mL of 0.068 M HCl is added to 1.000 L of the original buffer? (assume no volume change)
Show/Hide Answer
- pH = 5.222
- pH = 5.184
Refer to Section 7.1: Acid-Base Buffers (1).
Strategy Map
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Check out the strategy map.
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| Strategy Map Steps |
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| 1. Write out the ionization reaction for acetic acid (CH3COOH (aq)). |
2. Create an ICE (initial, change, equilibrium) table for the initial buffer solution.
Show/Hide HintPlug in the provided concentrations and solve for the concentration change using your Ka expression. |
3. Solve for hydronium concentration and calculate pH.
Show/Hide HintKeep track of your significant figures. |
4. Calculate the new species concentrations after adding acid.
Show/Hide HintAdding acid will cause 1 species to increase in concentration and 1 species to decrease in concentration. Make sure you correctly identify the direction of change. Show/Hide HintCalculate the concentration of the HCl in the 1.00 L of the buffer solution. |
5. Create a second ICE table using the new concentrations.
Show/Hide HintYour final concentrations from your first ICE table are your initial concentrations. |
6. Solve for hydronium concentration and calculate pH.
Show/Hide HintThink about how you expect the pH to change after adding strong acid. |
Solution
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Check out this solution.
Show/Hide Solution
a. What is the pH of the solution?
[latex]\mathrm{CH}_3\mathrm{COOH}\,\mathrm{(aq)}+\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\rightleftharpoons\mathrm{CH}_3\mathrm{COO}^-\mathrm{(aq)}+\mathrm{H}_3\mathrm{O}^+\mathrm{(aq)}[/latex]
| State | [CH3COOH] | [CH3COOH–] | [H3O+] |
|---|---|---|---|
| I | 0.200 | 0.600 | ~0 |
| C | -x | +x | +x |
| E | 0.200 − x | 0.600 + x | x |
| Assume x << 0.200 M | 0.200 | 0.600 | x |
Given: [latex]\mathrm{K}_\mathrm{a}=1.80\times10^{-5}[/latex]
[latex]\begin{aligned} 1.80\times10^{-5}&=\frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}_3\mathrm{O}^+]}{[\mathrm{CH}_3\mathrm{COOH}]} \\ 1.80\times10^{-5}&=\frac{\mathrm{0.600\mathrm{x}}}{0.200}\\ (1.80\times 10^{-5})(0.200)&=\mathrm{0.600\mathrm{x}}\\ \mathrm{x}&=\frac{\mathrm{1.80\times 10^{-5}(0.200)}}{(0.600)}\\ \mathrm{x}&=[\mathrm{H}_3\mathrm{O}^+]=6.00\times 10^{-6}\\\\ \mathrm{pH}=-\log{[\mathrm{H_3\mathrm{O}^+]}}&=-\log{(6.00\times10^{-6})}=5.222 \end{aligned}[/latex]
Answer: pH = 5.222
b. What is the pH of a solution that results when 0.500 mL of 0.068 M HCl is added to 1.000 L of the original buffer? (assume no volume change)
HCl (aq) initial concentration in the buffer:
[latex]\begin{gathered} =\frac{\frac{1\mathrm{~L}}{1000 \mathrm{~mL}}(0.0680\mathrm{~M})}{1.00\mathrm{~L}}=0.0340 \mathrm{~M} \end{gathered}[/latex]
| State | [CH3COOH] | [CH3COOH–] | [H3O+] |
|---|---|---|---|
| I — Adding H3O+ (aq) | 0.200 | 0.600 | 0.034 |
| C — Consuming H3O+ (aq) | +0.034 | -0.034 | -0.034 |
| E — After Consumption H3O+ (aq) | 0.234 | 0.556 | 0 |
| State | [CH3COOH] | [CH3COOH–] | [H3O+] |
|---|---|---|---|
| I | 0.234 | 0.556 | 0 |
| C | -x | +x | +x |
| E | 0.234 − x | 0.556 + x | x |
| Assume x << 0.234 M | 0.234 | 0.556 | x |
[latex]\begin{aligned} 1.80\times 10^{-5}&=\frac{\mathrm{[\mathrm{CH}_3\mathrm{COO}^-]}\mathrm{[\mathrm{H}_3\mathrm{O}^+]}}{[\mathrm{CH}_3\mathrm{COOH}]} \\ 1.80\times 10^{-5}&=\frac{\mathrm{0.556\mathrm{x}}}{0.234} \\ (1.80\times 10^{-5})(0.234)&=\mathrm{0.556\mathrm{x}} \\ \mathrm{x}&=\frac{\mathrm{1.80\times 10^{-5}(0.234)}}{(0.556)} \\ \mathrm{x}&=[\mathrm{H}_3\mathrm{O}^+]=7.58\times 10^{-6}\\\\ \mathrm{pH}=-\log[\mathrm{H}_3\mathrm{O}^+]&=-\log(7.58\times10^{-6})=5.120 \end{aligned}[/latex]
Answer: pH = 5.120
Guided Solution
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The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.
Show/Hide Guided Solution
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This question is a calculation problem that requires you to evaluate a buffer system and calculate the pH before and after a disturbance using ICE tables.
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A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 × 10−5 as Ka for acetic acid.
Show/Hide Think About This!To answer part a, we need to use an ICE table and the Ka expression to calculate the pH of the buffer solution. To answer part b, we need to recognize how the concentrations of the buffer species change when adding a strong acid and use another ICE table and the Ka expression to calculate the hydronium concentration and pH. |
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Write out the Ka reaction for acetic acid (C3COOH (aq)). This will help remind you which components are reactants and which are products for your ICE table.
Show/Hide Watch Out![latex]\mathrm{CH}_3\mathrm{COOH}\,\mathrm{(aq)}+\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\rightleftharpoons\mathrm{CH}_3\mathrm{COO}^-\mathrm{(aq)}+\mathrm{H}_3\mathrm{O}^+\mathrm{(aq)}[/latex] Recall that when an acid reacts with water, it will always produce hydronium (H3O+) as a product. This reaction is the equilibrium reaction taking place in the buffer solution. |
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For this problem, you will need to create 2 types of ICE tables.
Show/Hide Watch Out!You will need to create an ICE table for part a and another for part b. They will have different initial conditions. |
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Part b will have different initial conditions than part a because we are adding a strong acid, which disturbs the equilibrium. We must use a mini pre-table to determine the new concentrations.
Show/Hide Watch Out!Since we are adding a strong acid, we know we are essentially adding hydronium. To know if our acid or base is increasing, we identify which species will be reacting with the added hydronium. A buffer solution will react to try to consume the hydronium. The basic part of the buffer will react with the hydronium. Therefore, the [CH3COO–] will decrease and the [CH3COOH] will increase as the equilibrium shifts. |
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Set up your ICE table for part a.
Show/Hide Don’t Forget!ICE stands for initial, change, and equilibrium. We already know the initial concentrations of all species.
Show/Hide Don’t Forget!
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| Set up your initial ICE table for part b.
You are adding H3O+ (aq) to the original buffer solution. Make sure you think about how the original buffer solution will react to consume the added H3O+ (aq). Show/Hide Don’t Forget!
Show/Hide Don’t Forget!Calculate the concentration of H3O+ (aq) once HCl (aq) is added to the 1.00 L of buffer before the buffer reacts to consume it. Make assumptions to simplify the equilibrium concentration expressions. Show/Hide Don’t Forget!The buffer will react to take the added H3O+ (aq) concentration to 0. Then, the buffer will return to equilibrium through the Ka reaction. |
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| a. What is the pH of the solution?
Ionization reaction for acetic acid: [latex]\mathrm{CH}_3\mathrm{COOH}\,\mathrm{(aq)}+\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\rightleftharpoons\mathrm{CH}_3\mathrm{COO}^-\mathrm{(aq)}+\mathrm{H}_3\mathrm{O}^+\mathrm{(aq)}[/latex] Fill in your ICE table using initial concentrations:
We can assume the change is small compared to the initial buffer species concentrations. Given: Ka = 1.80 × 10-5 Plug the equilibrium concentrations from your ICE table in to the Ka expression, and solve for x: [latex]\begin{aligned} 1.80\times10^{-5}&=\frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}_3\mathrm{O}^+]}{[\mathrm{CH}_3\mathrm{COOH}]} \\ 1.80\times10^{-5}&=\frac{\mathrm{0.600\mathrm{x}}}{0.200}\\ (1.80\times 10^{-5})(0.200)&=\mathrm{0.600\mathrm{x}}\\ \mathrm{x}&=\frac{\mathrm{1.80\times 10^{-5}(0.200)}}{(0.600)}\\ \mathrm{x}&=[\mathrm{H}_3\mathrm{O}^+]=6.00\times 10^{-6} \end{aligned}[/latex] Recall the relationship between hydronium and the pH of the solution. Plug in the calculated hydronium concentration and solve for the buffer solution pH. [latex]\mathrm{pH}=-\log{[\mathrm{H_3\mathrm{O}^+]}}=-\log{(6.00\times10^{-6})}=5.222[/latex] |
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| b. What is the pH of a solution that results when 0.500 mL of 0.068 M HCl is added to 1.000 L of the original buffer? (assume no volume change)
Since we are adding a strong acid, we know we are essentially adding hydronium. To know if our acid or base is increasing, we identify which species will be reacting with the added hydronium. The buffer solution will react to try to consume the hydronium. The basic part of the buffer will react with the hydronium to consume it. Therefore, the [CH3COO–] will decrease and the [CH3COOH] will increase as the equilibrium shifts. [latex]\mathrm{CH}_3\mathrm{COO}^-\mathrm{(aq)}+\mathrm{H}_3\mathrm{O}^+\mathrm{(aq)}\rightleftharpoons\mathrm{CH}_3\mathrm{COOH}\,\mathrm{(aq)}+\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}[/latex] We are adding 0.500 mL of 0.068 M HCl (aq) to 1.000 L of the original buffer (assuming no change in volume). The HCl (aq) is being diluted by the buffer, so you are using a dilution calculation to determine the added H3O+ concentration: [latex]\begin{gathered} =\frac{\frac{1\mathrm{~L}}{1000 \mathrm{~mL}}(0.0680\mathrm{~M})}{1.00\mathrm{~L}}=0.0340 \mathrm{~M} \end{gathered}[/latex]
[latex]\begin{aligned} 1.80\times 10^{-5}&=\frac{\mathrm{[\mathrm{CH}_3\mathrm{COO}^-]}\mathrm{[\mathrm{H}_3\mathrm{O}^+]}}{[\mathrm{CH}_3\mathrm{COOH}]} \\ 1.80\times 10^{-5}&=\frac{\mathrm{0.556\mathrm{x}}}{0.234} \\ (1.80\times 10^{-5})(0.234)&=\mathrm{0.556\mathrm{x}} \\ \mathrm{x}&=\frac{\mathrm{1.80\times 10^{-5}(0.234)}}{(0.556)} \\ \mathrm{x}&=[\mathrm{H}_3\mathrm{O}^+]=7.58\times 10^{-6} \end{aligned}[/latex] Recall the relationship between hydronium and the pH of the solution. Plug in and solve for the buffer solution pH: [latex]\mathrm{pH}=-\log[\mathrm{H}_3\mathrm{O}^+]=-\log(7.58\times10^{-6})=5.120[/latex] |
Check Your Work
You were given an acidic buffer solution, so both calculated pH values should be less than 7. Once HCl (aq) is added, we would expect that if there is any change, it will be even more acidic.
Does your answer make chemical sense?
Show/Hide Answer
A buffer solution is designed to resist changes in pH. The answer makes chemical sense because it demonstrates the buffer is serving its purpose.
In our initial calculation (part a), we calculate the pH of the buffer solution itself, which is acidic. In our second calculation (part b), we calculate the pH of the solution after a strong acid is added. This added acid acts as a disturbance, which the buffer solution will try to counteract. The basic species in solution reacted to consume the added acid, and the buffer solution then returned to equilibrium.
The pH has dropped, meaning it made the solution slightly more acidic. We can mathematically see the buffer doing its job as the change was very small. Without the buffer system, the pH drop would have been much more significant.
PASS Attribution
- LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (2).
- Question 7.E.7 from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (3) is adapted under a CC BY-NC-SA 4.0 license.
- Question 7.E.7 is adapted from question 14.E.6.19: Q14.6.1 from LibreTexts Chemistry 1e (OpenSTAX) (4), which is under a CC BY 4.0 license.
- Question 14.E.6.19: Q14.6.1 is question 101 from OpenStax Chemistry (5), which is under a CC BY 4.0 license. Access for free at https://openstax.org/books/chemistry/pages/1-introduction.
References
1. OpenStax. 7.1 Acid-Base Buffers. In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/07%3A_Buffers_Titrations_and_Solubility_Equilibria/7.01%3A_Acid-Base_Buffers.
2. Thompson Rivers University. TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2024. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520).
3. Thompson Rivers University. 7.E: Buffers, Titrations and Solubility Equilibria (Exercises). In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2024. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/07%3A_Buffers_Titrations_and_Solubility_Equilibria/7.E%3A_Buffers_Titrations_and_Solubility_Equilibria_(Exercises).
4. OpenStax. 14.E: Acid-Base Equilibria (Exercises). In Chemistry 1e (OpenSTAX); LibreTexts, 2023. https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/14%3A_Acid-Base_Equilibria/14.E%3A_Acid-Base_Equilibria_(Exercises).
5. Flowers, P.; Robinson, W. R.; Langley, R.; Theopold, K. Ch. 14 Exercises. In Chemistry; OpenStax, 2015. https://openstax.org/books/chemistry/pages/14-exercises.
