AP Chemistry Quiz: Strong Acids And Bases Ph Poh

What this quiz covers

This quiz focuses on Strong Acids And Bases Ph Poh, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 3.00 (correct answer)
2. 11.00
3. 14.00
4. 1.00
5. 4.00

Explanation: This question tests the skill of calculating pH and pOH of strong acids and bases. HCl is a strong acid, meaning it fully dissociates in water to produce one H⁺ ion per molecule, so the [H⁺] equals the concentration of HCl, which is 1.0×10^{-3} M. The pH is calculated as -log[H⁺], so pH = -log($1.0×10^{-3}$) = 3.00. This direct relationship holds because there is no equilibrium to consider for strong acids. A tempting distractor is 11.00, which might result from confusing pH with pOH or mistakenly calculating for a base. For strong acids and bases, the ion concentration directly corresponds to the solute concentration (adjusted for stoichiometry) before taking the negative logarithm.

Question 2

1. 1.0×10^-10
2. 1.0×10^-4 (correct answer)
3. 1.0×10^-14
4. 1.0×10^4
5. 1.0×10^-8

Explanation: This question tests the skill of pH and pOH of strong acids and bases. HNO₃ is a strong acid, which means it fully dissociates in water according to HNO₃ → H⁺ + NO₃⁻, producing one H⁺ ion per molecule. Therefore, the concentration of H⁺ is equal to the concentration of HNO₃, which is 1.0×10^{-4} M. The pH would be calculated as -log[H⁺] = -log($1.0×10^{-4}$) = 4.00, but the question asks for [H⁺] directly. A tempting distractor might be 1.0×10^{-14}, if someone confused it with the ion product of water, K_w. For strong acids, the concentration directly gives ion concentration before taking logs.

Question 3

1. 3.00 (correct answer)
2. 11.00
3. 1.00
4. 14.00
5. 7.00

Explanation: 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.

Question 4

1. 1.0\times10^{-10} (correct answer)
2. 1.0\times10^{-14}
3. 1.0\times10^{-7}
4. 1.0\times10^{-2}
5. 1.0\times10^{-12}

Explanation: This question tests pH and pOH of strong acids and bases. KOH is a strong base that completely dissociates as KOH → K⁺ + OH⁻, producing [OH⁻] = 1.0×10⁻² M from a 1.0×10⁻² M solution. To find [H⁺], we use the water equilibrium constant: Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ at 25°C, so [H⁺] = Kw/[OH⁻] = (1.0×10⁻¹⁴)/(1.0×10⁻²) = 1.0×10⁻¹² M. A common mistake is using the OH⁻ concentration as the H⁺ concentration (choice B), forgetting they are inversely related. For strong bases, calculate [H⁺] by dividing Kw by [OH⁻].

Question 5

1. 1.30 (correct answer)
2. 12.70
3. 2.70
4. 1.70
5. 13.30

Explanation: 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.

Question 6

1. 6.0×10^-8
2. 6.0×10^-4
3. 1.0×10^-14
4. 1.2×10^-3 (correct answer)
5. 3.0×10^-4

Explanation: 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.

Question 7

1. 13.00 (correct answer)
2. 1.00
3. 12.00
4. 4.00
5. 10.00

Explanation: 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.

Question 8

1. 3.00 (correct answer)
2. 11.00
3. 1.00
4. 2.00
5. 13.00

Explanation: 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.

Question 9

1. 10.60
2. 3.40 (correct answer)
3. 2.70
4. 11.30
5. 3.70

Explanation: 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^{-4} = 4.0×10^{-4} M. The pOH is then -log[OH⁻] = -log($4.0×10^{-4}$) ≈ 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.

Question 10

1. 0.52 (correct answer)
2. 13.48
3. 1.48
4. 0.48
5. 1.52

Explanation: 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^{-1} M. Using log(3.0) ≈ 0.48, pOH = -log($3.0×10^{-1}$) = 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.

Question 11

1. 12.00 (correct answer)
2. 2.00
3. 14.00
4. 7.00
5. 1.00

Explanation: This question tests the skill of pH and pOH of strong acids and bases. HBr is a strong acid, which means it fully dissociates in water according to HBr → H⁺ + Br⁻, producing one H⁺ ion per molecule. Therefore, the concentration of H⁺ is equal to the concentration of HBr, which is 1.0×10^{-2} M. The pH is -log($10^{-2}$) = 2.00, and pOH = 14 - pH = 12.00. A tempting distractor might be 2.00, if someone calculated pH instead of pOH. For strong acids, the concentration directly gives ion concentration before taking logs.

Question 12

1. 13.00
2. 3.00 (correct answer)
3. 11.00
4. 1.00
5. 7.00

Explanation: This question tests the skill of pH and pOH of strong acids and bases. HClO$_4$ is a strong acid, which means it fully dissociates in water according to $\text{HClO}_4 \rightarrow \text{H}^{+} + \text{ClO}_4^{-}$, producing one H$^+$ ion per molecule. Therefore, the concentration of H$^+$ is equal to the concentration of HClO$_4$, which is $1.0 \times 10^{-3} \, \text{M}$. The pH is calculated as $-\log[\text{H}^{+}] = -\log(10^{-3}) = 3.00$. A tempting distractor might be 11.00, if someone calculated pOH instead of pH. For strong acids, the concentration directly gives ion concentration before taking logs.

Question 13

1. 1.0\times10^{-2}
2. 2.0\times10^{-2} (correct answer)
3. 5.0\times10^{-3}
4. 1.0\times10^{-12}
5. 2.0\times10^{-12}

Explanation: This question tests pH and pOH of strong acids and bases. Ba(OH)₂ is a strong base that completely dissociates according to Ba(OH)₂ → Ba²⁺ + 2OH⁻, releasing 2 moles of OH⁻ per mole of Ba(OH)₂. From a 0.010 M Ba(OH)₂ solution, the hydroxide concentration is [OH⁻] = 2 × 0.010 M = 0.020 M = 2.0×10⁻² M. A common mistake is forgetting the stoichiometric coefficient and using [OH⁻] = 1.0×10⁻² M (choice A), which ignores that each formula unit produces two hydroxide ions. For polyprotic bases, always multiply the base concentration by the number of OH⁻ groups in the formula.

Question 14

1. 13.70
2. 12.70 (correct answer)
3. 1.30
4. 0.70
5. 13.30

Explanation: 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^{-2} M. The pOH is -log($5×10^{-2}$) = 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.

Question 15

1. 11.30 (correct answer)
2. 2.70
3. 12.70
4. 3.30
5. 10.70

Explanation: 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.

Question 16

1. 1.0\times10^{-12}
2. 2.0\times10^{-2}
3. 1.0\times10^{-2} (correct answer)
4. 1.0\times10^{2}
5. 2.0\times10^{-12}

Explanation: 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.

Question 17

1. 14.00
2. 4.00
3. 2.00 (correct answer)
4. 12.00
5. 1.00

Explanation: 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^{-2} M. The pOH is -log($1.0×10^{-2}$) = 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.

Question 18

1. 0.020
2. 0.040 (correct answer)
3. 0.002
4. 0.200
5. 0.010

Explanation: This question tests the skill of pH and pOH of strong acids and bases. Ca(OH)$_2$ is a strong base, which means it fully dissociates in water according to $\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^{-}$, producing two OH$^{-}$ ions per formula unit. Therefore, the concentration of OH$^{-}$ is twice the concentration of Ca(OH)$_2$, which is $2 \times 0.020 \ \text{M} = 0.040 \ \text{M}$. The pOH would be calculated as $-\log[\text{OH}^{-}] = -\log(0.040) \approx 1.40$, 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.

Question 19

1. 13.00
2. 11.00
3. 14.00
4. 3.00 (correct answer)
5. 1.00

Explanation: 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^{-3} M. The pOH is calculated as -log[OH⁻], so pOH = -log($10^{-3}$) = 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.

Question 20

1. 2.00
2. 13.00
3. 11.00
4. 1.00
5. 3.00 (correct answer)

Explanation: 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^{-3} M. The pH is calculated as -log[H⁺], so pH = -log($10^{-3}$) = 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.