Acid-Base Equilibrium: Calculate pH and pOH for Strong Acid Solution
Question
Calculate the pH and the pOH of the following solution at 25 °C for which the substances ionize completely:
0.000259 M HClO4
Show/Hide Answer
pH = 3.587; pOH = 10.413
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. Write out the complete ionization reaction. You are told in the question it ionizes completely; this is because HClO4 (aq) is a strong acid. |
| 2. Since it ionizes completely, the given concentration of the strong acid is equal to the concentration of the hydronium ion, [H3O+]. |
| 3. Use the equation that relates pH and [H3O+] to solve for pH. |
4. Recall how pH and pOH are related and use that information to solve for pOH.
Show/Hide HintAt 25°C, pKw = 14.000 = pH + pOH |
5. Plug in the value you calculated for pH in step 2 to solve for pOH.
Show/Hide HintMake sure to pay attention to your significant figures. |
Solution
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Check out this solution.
Show/Hide Solution
[latex]\begin{gathered} \mathrm{HClO}_4\mathrm{(aq)}+\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\rightleftharpoons\mathrm{H}_3\mathrm{O}^+\mathrm{(aq)}+\mathrm{ClO}_4^-\mathrm{(aq)} \end{gathered}[/latex]
Calculate the pH:
[latex]\begin{aligned} \mathrm{pH}&=-\log{([\mathrm{H}_3\mathrm{O}^+])}\\ \mathrm{pH}&=-\log{(0.000259)}\\ \mathrm{pH}&=3.587 \end{aligned}[/latex]
Answer: pH = 3.587
Calculate the pOH:
[latex]\begin{aligned} \mathrm{pK}_\mathrm{w}&=14.000=\mathrm{pH}+\mathrm{pOH}\\ 14.000&=3.587+\mathrm{pOH}\\ 14.000-3.587&=\mathrm{pOH}\\ \mathrm{pOH}&=10.413 \end{aligned}[/latex]
Answer: pOH = 10.413
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 that tests your ability to find both the pH and pOH of a substance when given its concentration before ionization.
Show/Hide Resource
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| “Calculate the pH and the pOH of the following solution at 25°C for which the substances ionize completely: 0.000259 M HClO4”
We are told the initial concentration of the reactant as well as the temperature. Why is the temperature important? Show/Hide Think About This!We are told that the temperature the reaction occurs at is 25°C. Therefore, we know the pKw is 14.000. |
How can you use the provided initial concentration? Does it give the concentration of hydronium or hydroxide? Try writing out the complete ionization reaction.
Show/Hide Watch Out!The perchloric acid (HClO4) reacts with water to produce hydronium (H3O+) and perchlorate (ClO4–). [latex]\begin{gathered} \mathrm{HClO}_4\mathrm{(aq)}+\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\rightleftharpoons\mathrm{H}_3\mathrm{O}^+\mathrm{(aq)}+\mathrm{ClO}_4^-\mathrm{(aq)} \end{gathered}[/latex] |
Why is it important to know that the substance HClO4 ionizes completely?
Show/Hide Watch Out!Since it is completely ionized, we recognize that HClO4 (aq) is a strong acid and the initial acid concentration will equal the concentration of hydronium ion produced. |
Recall the equation that relates hydronium and pH.
Show/Hide Don’t Forget![latex]\begin{gathered} \mathrm{pH}=-\log{([\mathrm{H}_3\mathrm{O}^+])} \end{gathered}[/latex] Recall the equation that relates pH and pOH. Show/Hide Don’t Forget![latex]\begin{gathered} \mathrm{pK}_\mathrm{w}=14.000=\mathrm{pH}+\mathrm{pOH} \end{gathered}[/latex] How can we use these to find the information we want? Show/Hide Think About This!The first step is to use the hydronium concentration to solve for pH, then use pKw and pH to solve for pOH. |
Since you are working with logarithmic quantities, keeping track of your significant figures is important.
Show/Hide Don’t Forget![HClO4] = 0.000259 M, which has 3 significant figures When looking at a pH value, the digits before the decimal are not significant; so, to report a pH with 3 significant figures, you need 3 digits after the decimal. |
| Complete Solution |
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| The first step is to recognize that HClO4 (aq) is a strong acid that completely ionizes in an aqueous solution.
From this, we know that our provided concentration is also the concentration for hydronium. [latex]\begin{gathered} \mathrm{HClO}_4\mathrm{(aq)}+\mathrm{H}_2\mathrm{O}\,\mathrm{(l)}\rightleftharpoons\mathrm{H}_3\mathrm{O}^+\mathrm{(aq)}+\mathrm{ClO}_4^-\mathrm{(aq)} \end{gathered}[/latex] [latex]\begin{gathered} \,[\mathrm{HClO}_4]=0.000259\mathrm{M}=[\mathrm{H}_3\mathrm{O}^+] \end{gathered}[/latex] Show/Hide Don’t Forget!M which has 3 significant figures. Recall that when using logarithmic functions, your significant figures will be the number of digits after the decimal place in your calculation. Since our provided value is 0.000259, our calculation must have 3 significant digits after the decimal (__.123). |
| Recall the relationship between pH and hydronium. Plug your concentration value into the equation to solve for pH:
[latex]\begin{aligned} \mathrm{pH}&=-\log{([\mathrm{H}_3\mathrm{O}^+])}\\ \mathrm{pH}&=-\log{(0.000259)} \end{aligned}[/latex] Show/Hide Don’t Forget![HClO4] = 0.000259 M, which has 3 significant figures. Recall that when using logarithmic functions your significant figures will be the number of digits after the decimal place in your calculation. Since our provided value is 0.000259, our calculation must have 3 significant digits after the decimal (__.123). [latex]\begin{gathered} \mathrm{pH}=3.587 \end{gathered}[/latex] Answer: pH = 3.587 |
Recall the equation that relates pH and pOH.
Show/Hide Don’t Forget!We know the value of pKw since we are told the substance is at the standard temperature of 25 °C. Since our given concentration has 3 significant figures our value of pKw also has at least 3 significant figures. [latex]\begin{gathered} \mathrm{pK}_\mathrm{w}=14.000=\mathrm{pH}+\mathrm{pOH} \end{gathered}[/latex] Plug in the value calculated above for pH and rearrange to solve for pOH. [latex]\begin{aligned} 14.000&=3.587+\mathrm{pOH}\\ 14.000-3.587&=\mathrm{pOH}\\ \mathrm{pOH}&=10.413 \end{aligned}[/latex] Answer: pOH = 10.413 |
Check Your Work
If our calculation shows the final pressure as unchanged or less than the initial pressure, an error likely occurred in the calculation process.
Does your answer make chemical sense?
Show/Hide Answer
HClO4 (aq) is a strong acid. Therefore, we would expect the pH to be less than 7 and the pOH to be greater than 7. The pH and pOH should also add up to 14 since we are working at 25oC.
[latex]\begin{aligned} \mathrm{pH}&=3.587\\ \mathrm{pOH}&=10.413\\ \mathrm{pH}&=\mathrm{pOH}=3.587+10.413=14.000 \end{aligned}[/latex]
pH and pOH are scales used for measuring how acidic or basic a solution is. The higher the concentration of hydronium and the lower concentration of hydroxide, the more acidic the solution is. The higher the concentration of hydroxide and the lower the concentration of hydronium, the more basic the solution is.
In this problem, the substance being ionized was an acid, meaning the solution would produce more hydronium and be acidic. On the pH scale (which we are currently using), the lower the pH and the higher the pOH, the higher the acidity; this makes sense since our calculated values represented an acidic solution with this pattern.
PASS Attribution
- Question 6.E.12 (a) from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (2) is used under a CC BY-NC-SA 4.0 license.
- Question 6.E.12(a) is question 14.E.2.7: Q14.2.5(a) from LibreTexts Chemistry 1e (OpenSTAX) (3), which is under a CC BY 4.0 license.
- Question 14.E.2.7: Q14.2.5 (a) is question 19(a) from OpenStax Chemistry (4), 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); LireTexts, 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. 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.
