HI is a strong acid (Ka = 3.2 × 109). Calculate [H+], [OH−], pH and pOH of a 7.1 × 10−2 M HI solution.

[H + ] = 7.1 × 10 −2 M [OH − ] = 1.4 × 10 −13 M pH = 1.15 pOH = 12.85

HI is a strong acid (K a = 3.2 × 10 9 ). Calculate [H + ], [OH − ], pH and pOH of a 7.1 × 10 −2 M HI solution.

Hydroiodic acid is a strong acid and that's shown by the fact that our Ka is much greater than 1, it's 3.2 times 10 to the 9. This means that it completely ionizes into H+ ion and iodide ion. Now, here we need to calculate the H+ ion concentration, the hydroxide ion concentration, pH and pOH of a 7.1 times 10 to the negative 2 molar of hydroiodic acid solution. Alright, so let's do the first one that's easiest, H+ion. Remember, for a strong acid the H+ion concentration is equal to the molarity of the acid. Here we'd say that H plus is simply going to be the 7.1 times 10 to the negative 2 molar. So that would be our answer there for H+. Next, let's find Oh minus. Well, here we're going to say that k w equals h+timesoh minus. The temperature is not told to us, so we assume that it's 25 degrees Celsius and therefore k w is 1.0 times 10 to the negative 14. We plug in what we just found for h plus which is 7.1 times 10 to the negative two, times the concentration of o h minus. Now divide out the 7.1 times 10 to the negative 2 on both sides. When we do that we'll get our Oh- concentration, which comes out to be 1.4 times 10 to the negative 13 molar. Now that we know H+ ion and Oh minus ion, we can find pH and pOH. Remember that pH is just the negative log of H+. So it'd be negative log of 7.1 times 10 to the negative 2. This would give us a pH of 1.15. Finally, we could find pOH 1 of 2 ways. We could say that pH+pOH equals 14, plugging what we found from pH, subtracting it from 14 would give us pOH, or you could just say that pOH equals the negative log of Oh- and then plug in what we just found for Oh- which is 1.4 times 10 to the negative 13. Either way we'll get our pOH as equaling 12.85. So these would be each of the values that we needed to find for our hydroiodic solution.

HI is a strong acid (K a = 3.2 × 10 9 ). Calculate [H + ], [OH − ], pH and pOH of a 7.1 × 10 −2 M HI solution.

[H + ] = 7.1 × 10 −2 M [OH − ] = 1.4 × 10 −13 M pH = 1.15 pOH = 12.85

[H + ] = 7.1 × 10 −2 M [OH − ] = 1.4 × 10 −13 M pH = 8.50 pOH = 5.50

[H + ] = 7.1 × 10 −2 M [OH − ] = 7.1 × 10 −16 M pH = 1.15 pOH = 15.15

[H + ] = 7.1 × 10 −2 M [OH − ] = 7.1 × 10 −12 M pH = 1.15 pOH = 11.15

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1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
3. Ionic Compounds2h 18m
4. Molecular Compounds2h 18m
5. Classification & Balancing of Chemical Reactions3h 17m
6. Chemical Reactions & Quantities2h 27m
7. Energy, Rate and Equilibrium3h 46m
8. Gases, Liquids and Solids3h 25m
9. Solutions4h 10m
10. Acids and Bases3h 29m
11. Nuclear Chemistry56m
BONUS: Lab Techniques and Procedures1h 38m
BONUS: Mathematical Operations and Functions47m
12. Introduction to Organic Chemistry1h 34m
13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
14. Compounds with Oxygen or Sulfur1h 6m
15. Aldehydes and Ketones1h 1m
16. Carboxylic Acids and Their Derivatives1h 11m
17. Amines39m
18. Amino Acids and Proteins1h 51m
19. Enzymes1h 37m
20. Carbohydrates1h 41m
21. The Generation of Biochemical Energy2h 8m
22. Carbohydrate Metabolism2h 22m
23. Lipids2h 26m
24. Lipid Metabolism1h 45m
25. Protein and Amino Acid Metabolism1h 37m
26. Nucleic Acids and Protein Synthesis2h 54m

10. Acids and Bases

pH of Strong Acids and Bases

GOB Chemistry

10. Acids and Bases

pH of Strong Acids and Bases

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Multiple Choice

HI is a strong acid (K_a = 3.2 × 10⁹). Calculate [H⁺], [OH⁻], pH and pOH of a 7.1 × 10⁻² M HI solution.

[H⁺] = 7.1 × 10⁻² M

[OH⁻] = 1.4 × 10⁻¹³ M

pH = 1.15

pOH = 12.85

[H⁺] = 7.1 × 10⁻² M

[OH⁻] = 1.4 × 10⁻¹³ M

pH = 8.50

pOH = 5.50

[H⁺] = 7.1 × 10⁻² M

[OH⁻] = 7.1 × 10⁻¹⁶ M

pH = 1.15

pOH = 15.15

[H⁺] = 7.1 × 10⁻² M

[OH⁻] = 7.1 × 10⁻¹² M

pH = 1.15

pOH = 11.15

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Verified step by step guidance

Identify that HI is a strong acid, which means it completely dissociates in water. Therefore, the concentration of H+ ions is equal to the initial concentration of HI, which is 7.1 × 10^{-2} M.

Use the relationship between [H+] and [OH-] in water at 25°C, which is given by the ion product of water: Kw = [H+][OH-] = 1.0 × 10^{-14}. Substitute [H+] = 7.1 × 10^{-2} M into this equation to solve for [OH-].

Calculate the pH of the solution using the formula pH = -log[H+]. Substitute the value of [H+] = 7.1 × 10^{-2} M into this formula to find the pH.

Calculate the pOH of the solution using the relationship pOH = 14 - pH. Use the pH value obtained in the previous step to find the pOH.

Verify the consistency of your results by checking that the sum of pH and pOH equals 14, which is a characteristic of aqueous solutions at 25°C.

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