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Strong Acids and Bases, pH, pOH - AP Chemistry

  1. A 0.0010 M0.0010\,\text{M} solution of Sr(OH)2(aq)_2(aq) is prepared at 25∘C25^\circ\text{C}. Assuming complete dissociation, what is the pH of the solution?
    Answer: 11.30 This question tests pH and pOH of strong acids and bases. Sr(OH)₂ is a strong base that completely dissociates as Sr(OH)₂ → Sr²⁺ + 2OH⁻, producing [OH⁻] = 2 × 0.0010 M = 0.0020 M from a 0.0010 M solution. The pOH = -log[OH⁻] = -log(2.0×10⁻³) = 3 - log(2) ≈ 3 - 0.30 = 2.70, and pH = 14.00 - pOH = 14.00 - 2.70 = 11.30. A common error is forgetting the coefficient 2, which would give pOH = 3.00 and pH = 11.00. For diprotic bases, always account for both OH⁻ ions when calculating ion concentration.
  2. A 0.020 M0.020\ \text{M} Ca(OH)2_2(aq) solution is prepared at 25∘C25^\circ\text{C}. Assuming complete dissociation, what is [OH−][\text{OH}^-] in the solution?
    Answer: 0.040 This question tests the skill of pH and pOH of strong acids and bases. Ca(OH)2_2 is a strong base, which means it fully dissociates in water according to Ca(OH)2→Ca2++2OH−\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^{-}, producing two OH−^{-} ions per formula unit. Therefore, the concentration of OH−^{-} is twice the concentration of Ca(OH)2_2, which is 2×0.020 M=0.040 M2 \times 0.020 \ \text{M} = 0.040 \ \text{M}. The pOH would be calculated as −log⁡[OH−]=−log⁡(0.040)≈1.40-\log[\text{OH}^{-}] = -\log(0.040) \approx 1.40, but the question asks for [OH−^{-}] directly. A tempting distractor might be 0.020, if someone forgot to multiply by 2 for the two hydroxide ions. For strong bases, the concentration directly gives ion concentration before taking logs, adjusted for the number of ions per formula unit.
  3. A 1.0×10−2 M1.0\times10^{-2}\,\text{M} solution of HNO3(aq)_3(aq) is prepared at 25∘C25^\circ\text{C}. Assuming complete dissociation, what is [H+][\text{H}^+] in the solution?
    Answer: 1.0\times10^{-2} This question tests pH and pOH of strong acids and bases. HNO₃ is a strong acid that completely dissociates in water according to HNO₃ → H⁺ + NO₃⁻, producing one mole of H⁺ per mole of acid. For a 1.0×10⁻² M HNO₃ solution, the hydrogen ion concentration equals the acid concentration: [H⁺] = 1.0×10⁻² M. A common error is trying to calculate [OH⁻] instead, which would give 1.0×10⁻¹² M (choice A), but the question specifically asks for [H⁺]. For monoprotic strong acids, the H⁺ concentration always equals the initial acid concentration.
  4. A student prepares a 1.0×10−3 M1.0\times10^{-3}\,\text{M} solution of HCl(aq)(aq) at 25∘C25^\circ\text{C}. Assuming HCl dissociates completely in water, what is the pH of the solution?
    Answer: 3.00 This question tests pH and pOH of strong acids and bases. HCl is a strong acid that completely dissociates in water according to HCl → H⁺ + Cl⁻, meaning a 1.0×10⁻³ M HCl solution produces [H⁺] = 1.0×10⁻³ M. The pH is calculated as pH = -log[H⁺] = -log(1.0×10⁻³) = -(-3) = 3.00. A common mistake is confusing pH with pOH, which would give 11.00 (choice B), but pH directly uses [H⁺] concentration. For strong acids, the acid concentration equals the H⁺ concentration before taking the negative logarithm.
  5. A solution is made by dissolving NaOH(s)(s) in water to make 0.010 M0.010\ \text{M} NaOH(aq)(aq) at 25∘C25^\circ\text{C}. Assuming complete dissociation, what is the pOH of the solution?
    Answer: 2.00 This problem tests pH and pOH of strong acids and bases. NaOH is a strong base that completely dissociates: NaOH → Na⁺ + OH⁻, so [OH⁻] = [NaOH] = 0.010 M. To find pOH, we calculate pOH = -log[OH⁻] = -log(0.010) = -log(10⁻²) = 2.00. A common error would be to calculate pH instead of pOH, which would give 12.00 (choice A), but the question specifically asks for pOH. For strong bases, the base concentration directly gives the hydroxide ion concentration before taking logs.
  6. A student prepares a 1.0×10−3 M1.0\times10^{-3}\,\text{M} solution of HCl(aq)(aq) at 25∘C25^\circ\text{C}. Assuming HCl dissociates completely in water, what is the pH of the solution?​
    Answer: 3.00 This question tests pH and pOH of strong acids and bases. HCl is a strong acid that completely dissociates in water according to HCl → H⁺ + Cl⁻, meaning that [H⁺] equals the initial concentration of HCl. Since [HCl] = 1.0×10⁻³ M, we have [H⁺] = 1.0×10⁻³ M. The pH is calculated as pH = -log[H⁺] = -log(1.0×10⁻³) = 3.00. A common mistake would be to calculate pOH instead of pH, which would give 11.00 (choice B). For strong acids, remember that the acid concentration directly gives [H⁺] before taking the negative logarithm to find pH.
  7. A 1.0×10−4 M1.0\times10^{-4}\,\text{M} solution of NaOH(aq)(aq) is prepared at 25∘C25^\circ\text{C}. Assuming complete dissociation, what is the pH of the solution?
    Answer: 10.00 This question tests pH and pOH of strong acids and bases. NaOH is a strong base that completely dissociates as NaOH → Na⁺ + OH⁻, producing [OH⁻] = 1.0×10⁻⁴ M from a 1.0×10⁻⁴ M solution. To find pH, we first calculate pOH = -log[OH⁻] = -log(1.0×10⁻⁴) = 4.00, then use pH + pOH = 14.00 at 25°C to get pH = 14.00 - 4.00 = 10.00. A common mistake is calculating pOH = 4.00 and stopping there (choice A), forgetting to convert to pH. For strong bases, calculate pOH first, then subtract from 14 to find pH.
  8. A 2.0×10−4 M2.0\times10^{-4}\,\text{M} solution of Ba(OH)2(aq)_2(aq) is prepared in water at 25∘C25^\circ\text{C}. Ba(OH)2_2 is a strong base that completely dissociates. What is the pOH of the solution?
    Answer: 3.40 This question tests the skill of calculating pH and pOH of strong acids and bases. Ba(OH)₂ is a strong base that fully dissociates in water to produce two OH⁻ ions per formula unit, so [OH⁻] = 2 × 2.0×10−410^{-4} = 4.0×10−410^{-4} M. The pOH is then -log[OH⁻] = -log(4.0×10−44.0×10^{-4}) ≈ 3.40. This calculation assumes complete ionization without any equilibrium considerations. A tempting distractor is 10.60, which could come from calculating pH instead of pOH or forgetting the factor of 2. For strong acids and bases, the ion concentration directly gives the value before taking logs, remembering to multiply by the number of ions per molecule.
  9. A student prepares 0.0010 M0.0010\ \text{M} HCl(aq)(aq) by dissolving HCl gas in water at 25∘C25^\circ\text{C}. Assuming HCl is a strong acid that dissociates completely, what is the pH of the solution?
    Answer: 3.00 This question tests the skill of pH and pOH of strong acids and bases. HCl is a strong acid, which means it fully dissociates in water according to HCl → H⁺ + Cl⁻, producing one H⁺ ion per molecule. Therefore, the concentration of H⁺ is equal to the concentration of HCl, which is 0.0010 M or 10−310^{-3} M. The pH is calculated as -log[H⁺], so pH = -log(10−310^{-3}) = 3.00. A tempting distractor might be 2.00, if someone mistakenly thought the concentration was 0.01 M instead of 0.001 M. For strong acids, the concentration directly gives ion concentration before taking logs.
  10. A 3.0×10−1 M3.0\times10^{-1}\,\text{M} solution of LiOH(aq)(aq) is prepared at 25∘C25^\circ\text{C}. LiOH is a strong base that completely dissociates. What is the pOH of the solution? (Use log⁡(3.0)≈0.48\log(3.0)\approx0.48.)
    Answer: 0.52 This question tests the skill of calculating pH and pOH of strong acids and bases. LiOH is a strong base that fully dissociates to produce one OH⁻ ion per molecule, so [OH⁻] = 3.0×10−110^{-1} M. Using log(3.0) ≈ 0.48, pOH = -log(3.0×10−13.0×10^{-1}) = 1 - 0.48 = 0.52. This calculation assumes complete ionization. A tempting distractor is 13.48, perhaps from calculating pH = 14 - pOH incorrectly as 13.48. For strong acids and bases, concentration directly gives ion concentration before taking logs.
  11. A 8.0×10−3 M8.0\times10^{-3}\,\text{M} solution of HBr(aq)(aq) is prepared at 25∘C25^\circ\text{C}. Assuming HBr dissociates completely, what is the pH of the solution?​
    Answer: 2.10 This question tests pH and pOH of strong acids and bases. HBr is a strong acid that completely dissociates as HBr → H⁺ + Br⁻, so [H⁺] equals the initial HBr concentration. With [HBr] = 8.0×10⁻³ M, we have [H⁺] = 8.0×10⁻³ M. The pH = -log(8.0×10⁻³) = -log(0.008) = 2.10. A common error would be to calculate pOH instead, which would give 11.90 (choice B). For strong monoprotic acids, the acid concentration directly gives [H⁺], which is then used to calculate pH.
  12. A student dissolves Ba(OH)2_2(s)$ in water to make 0.0010 M0.0010\ \text{M} Ba(OH)2_2(aq)$ at 25∘C25^\circ\text{C}. Assuming complete dissociation, what is [OH−][\text{OH}^-]?
    Answer: 0.0020 This problem tests pH and pOH of strong acids and bases. Ba(OH)₂ is a strong base that completely dissociates: Ba(OH)₂ → Ba²⁺ + 2OH⁻, producing 2 hydroxide ions per formula unit. Therefore, [OH⁻] = 2 × 0.0010 M = 0.0020 M. A common error is forgetting the stoichiometric coefficient and using [OH⁻] = 0.0010 M (choice A), which would be true only for a 1:1 base like NaOH. For polyprotic bases, always multiply the base concentration by the number of OH⁻ ions in the formula.
  13. A 3.0×10−5 M3.0\times10^{-5}\,\text{M} solution of HNO3(aq)_3(aq) is prepared at 25∘C25^\circ\text{C}. Assuming HNO3_3 dissociates completely, what is [H+][\text{H}^+] in the solution?​
    Answer: 3.0×10−510^{-5} This question tests pH and pOH of strong acids and bases. HNO₃ is a strong acid that completely dissociates as HNO₃ → H⁺ + NO₃⁻, meaning [H⁺] equals the initial acid concentration. Since [HNO₃] = 3.0×10⁻⁵ M, we have [H⁺] = 3.0×10⁻⁵ M directly. A common mistake would be to double the concentration thinking there are multiple H⁺ ions (choice E), but HNO₃ produces only one H⁺ per molecule. For monoprotic strong acids, the H⁺ concentration always equals the initial acid concentration due to complete dissociation.
  14. A 6.0×10−4 M6.0\times10^{-4}\,\text{M} solution of Sr(OH)2(aq)_2(aq) is prepared at 25∘C25^\circ\text{C}. Assuming complete dissociation, what is [OH−][\text{OH}^-] in the solution?​
    Answer: 1.2×10−310^{-3} This question tests pH and pOH of strong acids and bases. Sr(OH)₂ is a strong base that completely dissociates as Sr(OH)₂ → Sr²⁺ + 2OH⁻, producing 2 moles of OH⁻ per mole of Sr(OH)₂. With [Sr(OH)₂] = 6.0×10⁻⁴ M, we calculate [OH⁻] = 2 × 6.0×10⁻⁴ = 1.2×10⁻³ M. A common mistake is forgetting the stoichiometric coefficient and using [OH⁻] = 6.0×10⁻⁴ M (choice A). For bases containing multiple hydroxide ions, multiply the base concentration by the number of OH⁻ ions in the formula.
  15. A solution contains 0.10 M0.10\ \text{M} KOH(aq)(aq) at 25∘C25^\circ\text{C}. Assuming complete dissociation, what is the pH of the solution?
    Answer: 13.00 This problem tests pH and pOH of strong acids and bases. KOH is a strong base that completely dissociates: KOH → K⁺ + OH⁻, so [OH⁻] = [KOH] = 0.10 M. First we find pOH = -log(0.10) = -log(10⁻¹) = 1.00, then pH = 14 - pOH = 14 - 1.00 = 13.00. A common mistake is to calculate pOH and report that as the answer (1.00, choice B), but the question asks for pH, not pOH. For strong bases, always check whether the question asks for pH or pOH to avoid this error.
  16. A student prepares 0.0010 M0.0010\ \text{M} HCl(aq)(aq) at 25∘C25^\circ\text{C}. Assuming HCl dissociates completely in water, what is the pH of the solution?
    Answer: 3.00 This problem tests pH and pOH of strong acids and bases. HCl is a strong acid that completely dissociates in water: HCl → H⁺ + Cl⁻, meaning [H⁺] = [HCl] = 0.0010 M. To find pH, we use pH = -log[H⁺] = -log(0.0010) = -log(10⁻³) = 3.00. A common mistake would be to calculate pOH instead of pH, which would give 11.00 (choice B), but the question specifically asks for pH. For strong acids, always remember that the acid concentration directly equals the H⁺ concentration before taking the negative logarithm.
  17. A solution is made by dissolving NaOH(s)(s) in water to form 1.0×10−3 M1.0\times10^{-3}\ \text{M} NaOH(aq)(aq) at 25∘C25^\circ\text{C}. Assuming NaOH is a strong base that dissociates completely, what is the pOH of the solution?
    Answer: 3.00 This question tests the skill of pH and pOH of strong acids and bases. NaOH is a strong base, which means it fully dissociates in water according to NaOH → Na⁺ + OH⁻, producing one OH⁻ ion per molecule. Therefore, the concentration of OH⁻ is equal to the concentration of NaOH, which is 1.0×10−310^{-3} M. The pOH is calculated as -log[OH⁻], so pOH = -log(10−310^{-3}) = 3.00. A tempting distractor might be 11.00, if someone calculated pH = 14 - pOH instead but for the wrong value. For strong bases, the concentration directly gives ion concentration before taking logs.
  18. A solution is prepared to be 0.050 M0.050\ \text{M} NaOH(aq)(aq) at 25∘C25^\circ\text{C}. Assuming complete dissociation, what is the pH of the solution? (Use log⁡(5)≈0.70\log(5)\approx0.70.)
    Answer: 12.70 This question tests the skill of pH and pOH of strong acids and bases. NaOH is a strong base, which means it fully dissociates in water according to NaOH → Na⁺ + OH⁻, producing one OH⁻ ion per molecule. Therefore, the concentration of OH⁻ is equal to the concentration of NaOH, which is 0.050 M or 5×10−210^{-2} M. The pOH is -log(5×10−25×10^{-2}) = 2 - log(5) ≈ 2 - 0.70 = 1.30, and pH = 14 - 1.30 = 12.70. A tempting distractor might be 1.30, if someone calculated pOH but reported it as pH. For strong bases, the concentration directly gives ion concentration before taking logs.
  19. A 1.0×10−2 M1.0\times10^{-2}\,\text{M} solution of NaOH(aq)(aq) is prepared at 25∘C25^\circ\text{C}. NaOH is a strong base that completely dissociates. What is the pOH of the solution?
    Answer: 2.00 This question tests the skill of calculating pH and pOH of strong acids and bases. NaOH is a strong base that fully dissociates to produce one OH⁻ ion per molecule, so [OH⁻] = 1.0×10−210^{-2} M. The pOH is -log(1.0×10−21.0×10^{-2}) = 2.00. This straightforward calculation assumes 100% ionization in water. A tempting distractor is 12.00, which might come from calculating pH = 14 - pOH incorrectly as 12. For strong acids and bases, concentration directly gives ion concentration before taking logs.
  20. A 0.050 M0.050\,\text{M} solution of HBr(aq)(aq) is prepared at 25∘C25^\circ\text{C}. Assuming complete dissociation, what is the pH of the solution? (Use log⁡(5)≈0.70\log(5)\approx0.70.)
    Answer: 1.30 This question tests pH and pOH of strong acids and bases. HBr is a strong acid that completely dissociates as HBr → H⁺ + Br⁻, producing [H⁺] = 0.050 M from a 0.050 M solution. The pH is calculated as pH = -log[H⁺] = -log(0.050) = -log(5×10⁻²) = 2 - log(5) ≈ 2 - 0.70 = 1.30. A common mistake is calculating pOH instead of pH, which would give pOH = 12.70 (choice B), but strong acids directly give H⁺ concentration for pH calculation. For strong acids, use the acid concentration as [H⁺] and calculate pH directly.

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