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AP Chemistry · Learn by Concept

AP Chemistry Help: Ph And Pk

Review real example questions for Ph And Pk in AP Chemistry.

Question 1 / 10

0 of 10 answered

For the weak acid HF in water, Ka=1.0×10−4K_a=1.0\times10^{-4}Ka​=1.0×10−4 at 25∘C25^\circ\text{C}25∘C. Which relationship between KaK_aKa​ and pKapK_apKa​ for HF is correct?

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All questions

Question 1

For the weak acid HF in water, Ka=1.0×10−4K_a=1.0\times10^{-4}Ka​=1.0×10−4 at 25∘C25^\circ\text{C}25∘C. Which relationship between KaK_aKa​ and pKapK_apKa​ for HF is correct?

  1. pKa=log⁡(Ka)pK_a=\log(K_a)pKa​=log(Ka​)
  2. pKa=−log⁡(Ka)pK_a=-\log(K_a)pKa​=−log(Ka​) (correct answer)
  3. pKa=1KapK_a=\dfrac{1}{K_a}pKa​=Ka​1​
  4. pKa=log⁡(1Ka)+1pK_a=\log\left(\dfrac{1}{K_a}\right)+1pKa​=log(Ka​1​)+1
  5. pKa=−log⁡([H+])pK_a=-\log([\text{H}^+])pKa​=−log([H+])

Explanation: This question assesses the skill of pH and pK. The pKa is the negative logarithm of the acid dissociation constant Ka, reflecting the strength of the acid where lower pKa indicates stronger acids due to greater dissociation. Similarly, pH is -log[H+], so both are logarithmic measures that allow relative comparisons without full calculations. For HF with Ka=1.0×10^{-4}, pKa=4, illustrating how pKa inversely relates to Ka's magnitude. A tempting distractor is choice A, which incorrectly uses pKa=log(Ka) without the negative sign, leading to positive values that don't align with typical pKa ranges for weak acids. Remember as a transferable strategy that lower pKa means stronger acid because it corresponds to a larger Ka.

Question 2

A weak acid HA has Ka=1.0×10−5K_a=1.0\times10^{-5}Ka​=1.0×10−5 at 25∘C25^\circ\text{C}25∘C. Which value is closest to pKapK_apKa​ for HA?

  1. 1
  2. 10
  3. 5.0×10−15.0\times10^{-1}5.0×10−1
  4. 1.0×10−51.0\times10^{-5}1.0×10−5
  5. 5 (correct answer)

Explanation: This question assesses the skill of pH and pK. The pKa is calculated as -log(Ka), transforming the exponential Ka into a linear scale for easier comparison of acid strengths. Lower pKa values indicate stronger acids due to the inverse logarithmic relationship. For Ka=1.0×10^{-5}, pKa=5, as -log(10^{-5})=5. A tempting distractor is choice D, which is the Ka value itself, confusing pKa with the dissociation constant rather than its logarithmic form. Remember as a transferable strategy that lower pKa means stronger acid because it corresponds to a larger Ka.

Question 3

A weak base BOH has pKb=4.0pK_b=4.0pKb​=4.0 at 25∘C25^\circ\text{C}25∘C. Which statement correctly describes the magnitude of KbK_bKb​?

  1. Kb=4.0×10−4K_b=4.0\times10^{-4}Kb​=4.0×10−4
  2. Kb=1.0×10−4K_b=1.0\times10^{-4}Kb​=1.0×10−4 (correct answer)
  3. Kb=1.0×104K_b=1.0\times10^{4}Kb​=1.0×104
  4. Kb=4.0×104K_b=4.0\times10^{4}Kb​=4.0×104
  5. Kb=1.0×10−10K_b=1.0\times10^{-10}Kb​=1.0×10−10

Explanation: This question assesses the skill of pH and pK. The pKb is -log(Kb), so Kb can be found as 10^{-pKb}, linking the logarithmic pKb to the base dissociation constant. Lower pKb indicates stronger bases with larger Kb values. For pKb=4.0, Kb=10^{-4}=1.0×10^{-4}. A tempting distractor is choice A, which uses 4.0×10^{-4} by mistakenly incorporating the digit 4 without proper logarithmic calculation. Remember as a transferable strategy that lower pKb means stronger base because it corresponds to a larger Kb.

Question 4

A weak base B has Kb=1.0×10−4K_b=1.0\times10^{-4}Kb​=1.0×10−4 at 25∘C25^\circ\text{C}25∘C. A student claims, “Because pKb=4.0pK_b=4.0pKb​=4.0, the pH of a solution of B must be 4.0.” Which statement best evaluates the claim?

  1. The claim is correct because pKbpK_bpKb​ equals the pH of any solution of that base.
  2. The claim is correct because pKbpK_bpKb​ equals −log⁡[OH−]-\log[\text{OH}^-]−log[OH−] for any solution.
  3. The claim is incorrect because pKbpK_bpKb​ is a property of the base, whereas pH depends on the extent of reaction and the solution concentration. (correct answer)
  4. The claim is incorrect because pKbpK_bpKb​ is defined as log⁡(Kb)\log(K_b)log(Kb​), not −log⁡(Kb)-\log(K_b)−log(Kb​).
  5. The claim is incorrect because pKbpK_bpKb​ must always be greater than 7.

Explanation: This question assesses the skill of pH and pK. The pKb is -log(Kb), a fixed property of the base indicating strength, while pH depends on [OH-] which varies with concentration and dissociation extent. Logarithmic relationships allow comparing strengths, but pH isn't equal to pKb. The claim equates pKb directly to pH, ignoring these factors. A tempting distractor is choice A, which incorrectly assumes pKb equals pH for any base solution, confusing the base's property with the solution's property. Remember as a transferable strategy that lower pKb means stronger base because it corresponds to a larger Kb, but pH requires concentration considerations.

Question 5

A weak acid HA has Ka=1.0×10−7K_a = 1.0\times 10^{-7}Ka​=1.0×10−7. Which relationship correctly gives pKapK_apKa​ for HA?

  1. pKa=−log⁡(Ka)pK_a = -\log(K_a)pKa​=−log(Ka​), so pKa=7.00pK_a = 7.00pKa​=7.00. (correct answer)
  2. pKa=log⁡(Ka)pK_a = \log(K_a)pKa​=log(Ka​), so pKa=−7.00pK_a = -7.00pKa​=−7.00.
  3. pKa=−log⁡([H+])pK_a = -\log([\text{H}^+])pKa​=−log([H+]), so pKapK_apKa​ equals the pH of the solution.
  4. pKa=−log⁡([HA])pK_a = -\log([\text{HA}])pKa​=−log([HA]), so pKapK_apKa​ depends on the initial acid concentration.
  5. pKa=1/KapK_a = 1/K_apKa​=1/Ka​, so pKa=1.0×107pK_a = 1.0\times 10^{7}pKa​=1.0×107.

Explanation: This question tests the mathematical relationship between pH and pK values. The pKa is defined as the negative logarithm (base 10) of the acid dissociation constant: pKa = -log(Ka). Given Ka = 1.0 × 10^-7, we apply this formula: pKa = -log(1.0 × 10^-7) = -(-7) = 7.00. This relationship is fundamental to acid-base chemistry and allows conversion between Ka and pKa values. Option B incorrectly omits the negative sign, while options C, D, and E confuse pKa with other quantities like pH or concentration. Remember: pKa = -log(Ka) is the defining relationship, just as pH = -log[H+].

Question 6

At 25∘C25^\circ\text{C}25∘C, two weak bases are compared: Base 1 has pKb=4.00pK_b = 4.00pKb​=4.00 and Base 2 has pKb=6.00pK_b = 6.00pKb​=6.00. Equal concentrations of each base are dissolved separately in water. Which statement is correct?

  1. Base 2 produces the more basic solution because its pKbpK_bpKb​ is larger.
  2. Base 1 produces the more basic solution because its pKbpK_bpKb​ is smaller. (correct answer)
  3. Both solutions have the same pH because both are weak bases.
  4. Base 1 is the weaker base because its pKbpK_bpKb​ is smaller.
  5. Both solutions must have pH=10\text{pH}=10pH=10 because pKbpK_bpKb​ values are given.

Explanation: This question tests understanding of pH and pK for comparing base strengths. Base 1 has pKb = 4.00 while Base 2 has pKb = 6.00, meaning Base 1 has the smaller pKb value. Since pKb = -log(Kb), a smaller pKb corresponds to a larger Kb value, indicating Base 1 is the stronger base. At equal concentrations, the stronger base (Base 1) will produce more OH- ions and create a more basic solution with higher pH. Option A incorrectly states that larger pKb produces a more basic solution, reversing the actual relationship. The key insight: smaller pKb means larger Kb, stronger base, and higher pH when comparing bases at the same concentration.

Question 7

Two separate 0.10 M0.10\,\text{M}0.10M aqueous solutions are prepared at the same temperature: Solution 1 contains weak acid HX with Ka=1.0×10−4K_a = 1.0\times 10^{-4}Ka​=1.0×10−4, and Solution 2 contains weak acid HY with Ka=1.0×10−6K_a = 1.0\times 10^{-6}Ka​=1.0×10−6. Which comparison is correct?

  1. Solution 1 has a higher pH because KaK_aKa​ is larger.
  2. Solution 2 has a lower pH because KaK_aKa​ is smaller.
  3. Both solutions have the same pH because both are 0.10 M0.10\,\text{M}0.10M.
  4. Both solutions have pH =4=4=4 because KaK_aKa​ values are powers of ten.
  5. Solution 1 has a lower pH because HX is the stronger acid (larger KaK_aKa​) at the same concentration. (correct answer)

Explanation: This question requires comparing pH and pK values for two weak acids. Since Ka represents the acid dissociation constant, a larger Ka value indicates a stronger acid that produces more H+ ions in solution. HX has Ka = 1.0 × 10^-4 while HY has Ka = 1.0 × 10^-6, making HX the stronger acid (its Ka is 100 times larger). At the same initial concentration, the stronger acid HX will produce more H+ ions and therefore have a lower pH than the weaker acid HY. Option A incorrectly states that larger Ka leads to higher pH, when the opposite is true. The key principle: larger Ka means stronger acid and lower pH when comparing acids at the same concentration.

Question 8

A 0.10 M0.10\,\text{M}0.10M solution of the weak acid HA has Ka=1.0×10−5K_a = 1.0\times 10^{-5}Ka​=1.0×10−5 (so pKa=5.00pK_a = 5.00pKa​=5.00). Which statement best relates the value of pKapK_apKa​ to the acidity of the solution?

  1. Because pKa=5.00pK_a = 5.00pKa​=5.00, the solution must have pH=5.00\text{pH} = 5.00pH=5.00.
  2. A smaller pKapK_apKa​ would indicate a stronger acid and thus a more acidic solution (lower pH), assuming the same initial concentration. (correct answer)
  3. A larger pKapK_apKa​ indicates a stronger acid and thus a lower pH, assuming the same initial concentration.
  4. pKapK_apKa​ is the negative logarithm of the initial acid concentration, so changing concentration changes pKapK_apKa​.
  5. Because KaK_aKa​ is less than 1, HA is a strong acid and the solution will have a very low pH.

Explanation: This question tests understanding of pH and pK relationships. The pKa value represents the negative logarithm of the acid dissociation constant (Ka), where pKa = -log(Ka). A smaller pKa value corresponds to a larger Ka value, which indicates a stronger acid that dissociates more completely in water. Since stronger acids produce more H+ ions at the same concentration, they result in lower pH values (more acidic solutions). The incorrect option C reverses this relationship by claiming larger pKa means stronger acid, when actually larger pKa means smaller Ka and therefore weaker acid. Remember: lower pKa means stronger acid and lower pH at the same concentration.

Question 9

A student states: “Because a solution has pKa=5.0K_a=5.0Ka​=5.0, its pH must be 5.0.” The student is referring to a 0.10 M0.10\,\text{M}0.10M aqueous solution of a weak monoprotic acid HA with pKa=5.0K_a=5.0Ka​=5.0 at 25∘C25^\circ\text{C}25∘C. Which statement best evaluates the student’s claim?​

  1. The claim is correct because pKaK_aKa​ is defined as the pH of the acid solution.
  2. The claim is incorrect because pKaK_aKa​ is a measure of acid strength, not the actual pH of a particular solution. (correct answer)
  3. The claim is correct because pKaK_aKa​ equals −log⁡[H+]-\log[\text{H}^+]−log[H+] for any acid solution.
  4. The claim is incorrect because pKaK_aKa​ must always be greater than 7 for a weak acid.
  5. The claim is correct because pKaK_aKa​ depends only on the initial concentration of HA.

Explanation: This question addresses a common misconception about pH and pK values. The pKa is a property of the acid itself, representing its strength through the relationship pKa = -log(Ka). The pH of a solution depends on both the acid's strength (Ka or pKa) and its concentration, requiring equilibrium calculations. For a 0.10 M solution of an acid with pKa = 5.0, the actual pH will be less than 5.0 (typically around 3.0) because the acid only partially dissociates. Choice A incorrectly defines pKa as the pH of the solution, confusing an intrinsic acid property with a solution property. Remember: pKa is a constant for a given acid at a specific temperature, while pH varies with concentration.

Question 10

Two weak bases, NH3\text{NH}_3NH3​ and B, are each dissolved separately in water to make solutions of the same initial concentration at 25∘C25^\circ\text{C}25∘C. The bases have KbK_bKb​ values Kb(NH3)=1.0×10−5K_b(\text{NH}_3)=1.0\times10^{-5}Kb​(NH3​)=1.0×10−5 and Kb(B)=1.0×10−3K_b(\text{B})=1.0\times10^{-3}Kb​(B)=1.0×10−3. Which statement correctly compares the pKbpK_bpKb​ values?

  1. pKb(NH3)>pKb(B)pK_b(\text{NH}_3) > pK_b(\text{B})pKb​(NH3​)>pKb​(B) because NH3\text{NH}_3NH3​ has the smaller KbK_bKb​. (correct answer)
  2. pKb(NH3)=pKb(B)pK_b(\text{NH}_3) = pK_b(\text{B})pKb​(NH3​)=pKb​(B) because both are weak bases.
  3. pKb(NH3)>pKb(B)pK_b(\text{NH}_3) > pK_b(\text{B})pKb​(NH3​)>pKb​(B) because NH3\text{NH}_3NH3​ has the larger KbK_bKb​.
  4. pKb(NH3)<pKb(B)pK_b(\text{NH}_3) < pK_b(\text{B})pKb​(NH3​)<pKb​(B) because NH3\text{NH}_3NH3​ has the smaller KbK_bKb​.
  5. pKb(NH3)<pKb(B)pK_b(\text{NH}_3) < pK_b(\text{B})pKb​(NH3​)<pKb​(B) because NH3\text{NH}_3NH3​ has the larger KbK_bKb​.

Explanation: This question assesses the skill of pH and pK. The pKb measures base strength logarithmically as -log(Kb), where a smaller pKb indicates a larger Kb and stronger base. For bases of the same concentration, relative pKb values allow comparison without calculating exact pOH or pH. Here, NH3 with smaller Kb has higher pKb than B, indicating it's weaker. A tempting distractor is choice A, which incorrectly states pKb(NH3) < pKb(B) despite NH3's smaller Kb, misunderstanding the inverse logarithmic relationship. Remember as a transferable strategy that lower pKb means stronger base because it corresponds to a larger Kb.