Acid-Base Equilibrium: Calculate Hydronium concentration, Hydroxide concentration, pH, pOH
Question
The ionization constant for water (Kw) is 2.90 × 10−14 at 40°C.
Calculate [H3O+], [OH−], pH, and pOH for pure water at 40°C.
Show/Hide Answer
- pH = pOH = 6.769
- [OH−] = [H3O+] = 1.70 × 10−7 M
Refer to Section 6.2: pH and pOH (1).
Strategy Map
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Check out the strategy map.
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| Strategy Map Steps |
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1. Recognize the question is about pure water at 40°C and gives the Kw.
Show/Hide HintSince we are looking at neutral water, recall that pH = pOH and [H3O+] = [OH–]. |
| Strategy 2: Solve for pH or pOH first
2. Recall how to solve for pKw given Kw. Use pKw to solve for pH and pOH. Use the relationship between pH and [H3O+] to solve for [H3O+] and [OH–]. Show/Hide Hint[latex]\begin{aligned} \mathrm{pK}_{\mathrm{w}}&=-\log\mathrm{K}_{\mathrm{w}}\\ \mathrm{pH}+\mathrm{pOH}&=\mathrm{pK}_{\mathrm{w}} \end{aligned}[/latex] |
| Strategy 2: Solve for [H3O+]/[OH–] first
2. Recall that Kw is the concentrations of H3O+ and OH– multiplied by each other. Use this relationship to solve for their concentrations. 3. Use the relationship between pH and [H3O+] to solve for pH and pOH. Show/Hide Hint[latex]\mathrm{K}_\mathrm{w}=[\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-][/latex] |
Solution
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Key equations:
[latex]\begin{aligned} \mathrm{pH}&=-\log[\mathrm{H}_3\mathrm{O}^+]\\ \mathrm{pOH}&=-\log[\mathrm{OH}^-] \end{aligned}[/latex]
There are 2 ways to solve this question.
Strategy 1: Solve for pH or pOH first.
[latex]\begin{aligned} \mathrm{pK}_\mathrm{w}&=-\log(\mathrm{K}_\mathrm{w})\\ \mathrm{pK}_{\mathrm{w}}&=-\log(2.90\times 10^{-14})\\ \mathrm{pK}_{\mathrm{w}}&=13.538\\\\ \mathrm{pK}_{\mathrm{w}}&=\mathrm{pH}+\mathrm{pOH}\\ \frac{13.538}{2}&=6.769=\mathrm{pH}=\mathrm{pOH}\\ \end{aligned}[/latex]
[latex]\begin{gathered} \,[\mathrm{H}_3\mathrm{O}^{+}]=10^{-\mathrm{pH}}\text{ and }[\mathrm{OH}^{-}]=10^{-\mathrm{pOH}}\\ 10^{-6.769}=[\mathrm{OH}^{-}]=[\mathrm{H}_3\mathrm{O}^{+}]=1.70\times 10^{-7}\mathrm{M} \end{gathered}[/latex]
Strategy 2: Solve for [H3O+] or [OH-] first.
[latex]\begin{aligned} \mathrm{K}_{\mathrm{w}}&=\left[\mathrm{H}_3\mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\\ 2.90 \times 10^{-14}&=\left[\mathrm{H}_3\mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\\ 2.90 \times 10^{-14}&=\left[\mathrm{OH}^{-}\text{or }\mathrm{H}_3\mathrm{O}^{+}\right]^2\\ \sqrt{2.90\times 10^{-14}}&=1.70 \times 10^{-7}\mathrm{M}=\left[\mathrm{H}_3\mathrm{O}^{+}\right]=[\mathrm{OH}] \end{aligned}[/latex]
[latex]\begin{gathered} \mathrm{pH}=-\log\left[\mathrm{H}_3\mathrm{O}^{+}\right]\text{ and }\mathrm{pOH}=-\log\left[\mathrm{OH}^{-}\right]\\ -\log\left(1.70\times 10^{-7}\right)=\mathrm{pH}=\mathrm{pOH}=6.769 \end{gathered}[/latex]
Answer:
- pH = pOH = 6.769
- [OH−] = [H3O+] = 1.70 × 10−7 M
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.
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| Guided Solution Ideas |
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This question is a calculation type problem where you must use your knowledge on relationships relating to pH and pOH.
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| What information is given in the question? What are you asked to find?
“The ionization constant for water (Kw) is 2.90 × 10−14 at 40 °C. Calculate [H3O+], [OH−], pH, and pOH for pure water at 40 °C.” Think Think About This!You are told that the Kw value is 2.90 × 10-14 this will be used in your calculations. You are also told that you are evaluating pure water. You are asked to find its [H3O+], [OH−], pH, and pOH. |
Recall these important relationships:
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What is the significance of pure water?
Show/Hide Think About This!Pure water is a contains only H3O+(aq) and OH–(aq) ions at equal concentrations. This means pOH and pH will be equal due to the concentrations of H3O+ and OH– being equal. |
Does it matter what you solve for first?
Think Think About This!Due to the relationships associated with pH and pOH, it does not matter which one you solve for first. There are multiple “pathways” to arrive at the same answers. |
| Complete Solution |
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| Pure water is a neutral substance. This means pOH and pH will be equal due to the concentrations of H3O+ and OH– being equal.
Recall what your significant figures will be after the decimal place when working with logarithms. Since there are 3 significant figures in 2.90 × 10-14, your answer will have 3 significant digits after the decimal place. Recall that the liquid is pure water and how that plays a role in the calculation. |
| Approach 1: Solve for pH and pOH first.
Recall the relationship between pKw and Kw: [latex]\mathrm{pK}_{\mathrm{w}}=-\log(\mathrm{K}_{\mathrm{w}})[/latex] Plug in your value for Kw: [latex]\mathrm{pK}_{\mathrm{w}}=-\log(2.90 \times 10^{-14})[/latex] Solve the logarithm: [latex]\mathrm{pK}_{\mathrm{w}}=13.538[/latex] Now, use pKw to solve for pH and pOH: [latex]\mathrm{pK}_{\mathrm{w}}=\mathrm{pH}+\mathrm{pOH}[/latex] Since your pH and pOH values are the same, divide your pKw by 2 to get their value: [latex]\begin{gathered} \frac{13.538}{2}=6.769=\mathrm{pH}=\mathrm{pOH} \end{gathered}[/latex] Now, recall the relationship between pH and H3O+, pOH, and OH–: [latex]\,[\mathrm{H}_3 \mathrm{O}^{+}]=10^{-\mathrm{pH}}\text{ and }[\mathrm{OH}^{-}]=10^{-\mathrm{pOH}}[/latex] Plug in your pH value to solve for your concentrations: [latex]10^{-6.769}=[\mathrm{OH}^{-}]=[\mathrm{H}_3\mathrm{O}^{+}]=1.70\times 10^{-7}\mathrm{M}[/latex] Answer:
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| Approach 2: Solve for [H3O+]/[OH-] first.
Recall the relationship between Kw and the concentrations of H3O+ and OH–: [latex]\mathrm{K}_{\mathrm{w}}=[\mathrm{H}_3\mathrm{O}^{+}][\mathrm{OH}^{-}][/latex] Plug in your value for Kw: [latex]2.90\times10^{-14}=[\mathrm{H}_3\mathrm{O}^{+}][\mathrm{OH}^{-}][/latex] Recall that the values for OH– and H3O+ are equal, so the value is squared: [latex]2.90\times 10^{-14}=[\mathrm{OH}^{-}]\text{ or }[\mathrm{H}_3\mathrm{O}^{+}]^2[/latex] Take the square root of both sides (squaring your Kw value) to solve for your concentrations: [latex]\sqrt{2.90\times 10^{-14}}=1.70\times 10^{-7}\mathrm{M}=[\mathrm{H}_2\mathrm{O}^{+}]=[\mathrm{OH}^{-}][/latex] Recall the relationship between your concentrations and pH and pOH: [latex]\mathrm{pH}=-\log(\mathrm{H}_2 \mathrm{O}^{+})\text{ and }\mathrm{pOH}=-\log(\mathrm{OH}^{-})[/latex] Plug in the value you calculated above and solve for pH and pOH: [latex]-\log(1.70\times 10^{-7})=\mathrm{pH}=\mathrm{pOH}=6.769[/latex] Answer:
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Check Your Work
Summary of what we would expect based on the related chemistry theory
Show/Hide Watch Out!
Make sure you have used the provided Kw value for the given temperature (40 ℃) and tracked your significant figures.
Kw = 2.90 × 10−14 at 40 °C
If pH = 6.300, then [H3O+] = 5.01; each term has 3 significant figures.
Does your answer make chemical sense?
Show/Hide Answer
Pure water undergoes autoionization to a small extent, producing small concentrations of H3O+ (aq) and OH– (aq) ions. Their mole ratio is 1:1, and pure water will always have equal concentrations of H3O+ and OH– because it only contains those 2 ions from the autoionization reaction. This means the pH and pOH will also always be equal.
However, these values change due to temperature, since equilibrium constants, including Kw, are temperature dependent. As temperatures increase and decrease, the value of Kw will change.
All of these properties are related and explained through their various relationships. For this reason, we can calculate all other values when we are only provided with 1.
Autoionization of water reaction:
[latex]\begin{gathered} \mathrm{H}_2\mathrm{O}(\mathrm{l})+\mathrm{H}_2\mathrm{O}(\mathrm{l})\rightleftharpoons\mathrm{H}_3\mathrm{O}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\\ \mathrm{K}_{\mathrm{w}}=[\mathrm{H}_3\mathrm{O}^{+}][\mathrm{OH}^{-}] \end{gathered}[/latex]
At 25℃, Kw is 1.00 × 10-14 and the pH and pOH of pure water is 7.000.
As the temperature increases to 40 ℃, the value of Kw is larger which means the ion concentrations are also larger and the pH and pOH have changed accordingly.
PASS Attribution
- LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (2)
- Question 6.E.11 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 6.E.11 is question 14.E.2.3: Q14.2.2 from LibreTexts Chemistry 1e (OpenSTAX) (4), which is under a CC BY 4.0 license.
- Question 14.E.2.3: Q14.2.2 is question 16 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. 6.2 pH and pOH. 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)/06%3A_Acid-Base_Equilibrium/6.02%3A_pH_and_pOH.
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. OpenStax. 6.E: Acid-Base Equilibrium (Exercises). 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)/06%3A_Acid-Base_Equilibrium/6.E%3A_Acid-Base_Equilibrium_(Exercises)
4. OpenStax. 14.E: Acid-Base Equilibria (Exercises). In Chemistry 1e (OpenSTAX); LibreTexts, 2022. 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.
