College Chemistry Quiz: Henderson Hasselbalch Equation

20 questions · exam conditions

0:00

Henderson Hasselbalch EquationQuestion 1 of 20

A phosphate buffer system contains 0.080 M $H_2PO_4^-$ and 0.120 M $HPO_4^{2-}$ . Given that $K_{a2}$ for phosphoric acid is $6.2 \times 10^{-8}$ , what is the pH of this buffer?

7.04

7.21

7.38

7.55

7.72

← Back to quizzes

College Chemistry Quiz

College Chemistry Quiz: Henderson Hasselbalch Equation

Practice Henderson Hasselbalch Equation in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Henderson Hasselbalch Equation, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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.

All questions

Question 1

A phosphate buffer system contains 0.080 M $H_2PO_4^-$ and 0.120 M $HPO_4^{2-}$ . Given that $K_{a2}$ for phosphoric acid is $6.2 \times 10^{-8}$ , what is the pH of this buffer?

7.04
7.21
7.38 (correct answer)
7.55
7.72

Explanation: When you encounter a buffer system question, you're dealing with the Henderson-Hasselbalch equation, which relates pH to the ratio of conjugate base to weak acid concentrations. For this phosphate buffer containing

H_2PO_4^-

(the weak acid) and

HPO_4^{2-}

(the conjugate base), you'll use:

pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)

First, calculate

pK_a

pK_a = -\log(6.2 \times 10^{-8}) = 7.21

Then substitute the concentrations:

pH = 7.21 + \log\left(\frac{0.120}{0.080}\right) = 7.21 + \log(1.5) = 7.21 + 0.18 = 7.39

This rounds to 7.38, confirming answer C is correct. Let's examine why the other answers are wrong: Answer A (7.04) would result from incorrectly flipping the concentration ratio, giving

7.21 + \log(0.80/1.20) = 7.21 - 0.18 = 7.03

. Answer B (7.21) represents just the

pK_a

value without accounting for the concentration ratio - this would only be correct if the concentrations were equal. Answer D (7.55) likely comes from calculation errors in either the

pK_a

determination or logarithm evaluation. Remember this key strategy: In Henderson-Hasselbalch problems, always identify which species is the weak acid and which is the conjugate base, then carefully set up your concentration ratio as [base]/[acid]. The pH will be higher than

pK_a

when the base concentration exceeds the acid concentration, as it does here.

Question 2

A buffer contains 0.15 M formic acid ( $HCOOH$ , $K_a = 1.8 \times 10^{-4}$ ) and 0.20 M sodium formate ( $HCOONa$ ). If 10.0 mL of 0.10 M $NaOH$ is added to 100.0 mL of this buffer, what is the new pH?

3.88
4.02 (correct answer)
4.16
4.30
4.44

Explanation: When you encounter a buffer problem with added strong base, you're dealing with acid-base equilibrium and the Henderson-Hasselbalch equation. The key is recognizing that the added

NaOH

will react with the weak acid component before affecting the buffer's pH significantly. First, calculate the moles of each component. In 100.0 mL of buffer: 0.015 mol

HCOOH

and 0.020 mol

HCOO^-

. The added

NaOH

contributes 0.0010 mol

OH^-

. The

OH^-

reacts completely with

HCOOH

HCOOH + OH^- → HCOO^- + H_2O

After reaction:

HCOOH

= 0.015 - 0.001 = 0.014 mol, and

HCOO^-

= 0.020 + 0.001 = 0.021 mol. The total volume is now 110.0 mL, so the new concentrations are [

HCOOH

] = 0.127 M and [

HCOO^-

] = 0.191 M. Using Henderson-Hasselbalch:

pH = pK_a + \log\frac{[A^-]}{[HA]} = 3.74 + \log\frac{0.191}{0.127} = 3.74 + 0.28 = 4.02

Choice A (3.88) likely results from forgetting to account for the volume change when calculating new concentrations. Choice C (4.16) and D (4.30) probably stem from calculation errors in the logarithm or incorrectly determining the moles after the neutralization reaction. Remember: in buffer problems with added strong acid or base, always perform the neutralization reaction first, then apply Henderson-Hasselbalch with the new concentrations. Don't forget to account for the volume change from the added solution.

Question 3

A buffer contains 0.040 M benzoic acid ( $C_6H_5COOH$ , $K_a = 6.3 \times 10^{-5}$ ) and 0.060 M sodium benzoate. After adding 5.0 mL of 0.20 M $HCl$ to 50.0 mL of this buffer, what is the final pH?

3.95
4.02 (correct answer)
4.10
4.20
4.35

Explanation: Buffer problems test your understanding of how acid-base equilibria respond to added strong acids or bases. When you see a buffer with added strong acid, think Henderson-Hasselbalch equation after accounting for the reaction between the strong acid and the buffer's base component. First, calculate the moles before reaction: benzoic acid = 0.040 M × 0.050 L = 0.002 mol, sodium benzoate = 0.060 M × 0.050 L = 0.003 mol, and added HCl = 0.20 M × 0.005 L = 0.001 mol. The added HCl reacts completely with benzoate ions:

C_6H_5COO^- + HCl → C_6H_5COOH + Cl^-

After this reaction, you have: benzoic acid = 0.002 + 0.001 = 0.003 mol, and benzoate = 0.003 - 0.001 = 0.002 mol. The total volume is now 55.0 mL, so the new concentrations are: [

C_6H_5COOH

] = 0.003 mol/0.055 L = 0.0545 M and [

C_6H_5COO^-

] = 0.002 mol/0.055 L = 0.0364 M. Using Henderson-Hasselbalch:

pH = pK_a + \log\frac{[A^-]}{[HA]} = -\log(6.3 × 10^{-5}) + \log\frac{0.0364}{0.0545} = 4.20 + (-0.18) = 4.02

Answer B (4.02) is correct. Answer A (3.95) likely results from calculation errors in the logarithm. Answer C (4.10) might come from using incorrect concentrations or forgetting the volume change. Answer D (4.20) represents the original buffer pH before adding HCl. Remember: always account for the stoichiometric reaction first, then apply Henderson-Hasselbalch with the final concentrations and total volume.

Question 4

An amino acid buffer system contains glycine in both its zwitterionic form ( $^+NH_3CH_2COO^-$ ) and its anionic form ( $NH_2CH_2COO^-$ ). If the $pK_a$ for the amino group is 9.60 and the buffer has a pH of 10.10, what is the ratio $\frac{[NH_2CH_2COO^-]}{[^+NH_3CH_2COO^-]}$ ?

1.3
2.2
3.2 (correct answer)
4.8
6.3

Explanation: When you encounter amino acid buffer problems, you're dealing with Henderson-Hasselbalch equilibrium between different ionization states. The key is identifying which forms are in equilibrium and applying the equation correctly. For this glycine buffer, you have the zwitterionic form (

^+NH_3CH_2COO^-

) losing a proton from its amino group to become the anionic form (

NH_2CH_2COO^-

). The equilibrium is:

^+NH_3CH_2COO^- \rightleftharpoons NH_2CH_2COO^- + H^+

Using the Henderson-Hasselbalch equation:

pH = pK_a + \log\frac{[A^-]}{[HA]}

Here, the deprotonated form (

NH_2CH_2COO^-

) is the base and the zwitterion is the acid. Substituting the values:

10.10 = 9.60 + \log\frac{[NH_2CH_2COO^-]}{[^+NH_3CH_2COO^-]}

0.50 = \log\frac{[NH_2CH_2COO^-]}{[^+NH_3CH_2COO^-]}

\frac{[NH_2CH_2COO^-]}{[^+NH_3CH_2COO^-]} = 10^{0.50} = 3.16 \approx 3.2

This confirms answer C is correct. Answer A (1.3) would result from incorrectly using

pH = 9.13

. Answer B (2.2) might come from calculation errors or using the wrong pKa value. Answer D (4.8) could result from confusing which form goes in the numerator versus denominator. Remember: when pH > pKa, the deprotonated (basic) form predominates. Always double-check which species is the acid and which is the base in your Henderson-Hasselbalch setup.

Question 5

A citrate buffer contains 0.08 M citric acid ( $H_3Cit$ ) and 0.12 M dihydrogen citrate ( $H_2Cit^-$ ). Given that $K_{a1} = 7.4 \times 10^{-4}$ for citric acid, what is the pH of this buffer?

2.95
3.13
3.31 (correct answer)
3.49
3.67

Explanation: When you encounter a buffer problem involving a weak acid and its conjugate base, you should immediately think of the Henderson-Hasselbalch equation:

pH = pK_a + \log\frac{[A^-]}{[HA]}

For this citrate buffer, you're dealing with the first ionization of citric acid (

H_3Cit

) forming dihydrogen citrate (

H_2Cit^-

). First, calculate

pK_{a1}

pK_{a1} = -\log(7.4 \times 10^{-4}) = 3.13

Now apply Henderson-Hasselbalch with

[H_2Cit^-] = 0.12

M and

[H_3Cit] = 0.08

pH = 3.13 + \log\frac{0.12}{0.08} = 3.13 + \log(1.5) = 3.13 + 0.18 = 3.31

Answer A (2.95) would result if you incorrectly flipped the ratio in the Henderson-Hasselbalch equation, giving

3.13 + \log(0.67) = 3.13 - 0.18 = 2.95

. Answer B (3.13) is simply the

pK_a

value, which would only be correct if the concentrations of acid and conjugate base were equal. Answer D (3.49) might result from calculation errors or using the wrong

K_a

value. The key study tip for buffer problems: always identify which acid-base pair you're working with (especially important for polyprotic acids like citric acid), calculate the

pK_a

first, then carefully apply Henderson-Hasselbalch with the conjugate base concentration in the numerator.

Question 6

An acetate buffer with pH 4.50 is diluted by adding pure water until the volume doubles. Assuming no other changes occur, what is the new pH of the diluted buffer?

4.20
4.35
4.50 (correct answer)
4.65
4.80

Explanation: This question tests your understanding of buffer behavior during dilution, a fundamental concept in acid-base chemistry. When you encounter buffer dilution problems, remember that buffers resist pH changes due to the equilibrium between a weak acid and its conjugate base. The Henderson-Hasselbalch equation governs buffer pH:

pH = pK_a + \log\frac{[A^-]}{[HA]}

. For an acetate buffer, this becomes

pH = pK_a + \log\frac{[acetate^-]}{[acetic\ acid]}

. The key insight is that when you dilute a buffer by adding water, you decrease the concentrations of both the weak acid and conjugate base by exactly the same factor. Since both concentrations are halved during this 2:1 dilution, their ratio

\frac{[A^-]}{[HA]}

remains unchanged. Because the ratio is constant, the log term stays the same, and the pH remains 4.50. Choice A (4.20) incorrectly assumes the pH decreases significantly, perhaps from confusing buffer dilution with simple acid dilution. Choice B (4.35) represents a smaller but still incorrect pH drop, possibly from applying weak acid dilution formulas inappropriately. Choice D (4.65) suggests the pH increases, which would violate buffer principles entirely. Study tip: Remember that buffer pH depends only on the ratio of conjugate base to weak acid concentrations, not their absolute values. Dilution affects both components equally, leaving the ratio—and thus the pH—unchanged. This is what makes buffers so effective at maintaining stable pH in biological systems.

Question 7

A buffer solution contains 0.15 M $HPO_4^{2-}$ and 0.10 M $PO_4^{3-}$ . If $K_{a3} = 4.2 \times 10^{-13}$ for the third ionization of phosphoric acid, what is the pH of this buffer?

11.85
12.03
12.21 (correct answer)
12.39
12.57

Explanation: When you encounter a buffer problem involving a weak acid and its conjugate base, you need to apply the Henderson-Hasselbalch equation:

pH = pK_a + \log\frac{[A^-]}{[HA]}

. Here, you're working with the third ionization step of phosphoric acid:

HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}

. In this equilibrium,

HPO_4^{2-}

acts as the weak acid and

PO_4^{3-}

is its conjugate base. First, calculate

pK_{a3}

pK_{a3} = -\log(4.2 \times 10^{-13}) = 12.38

Now apply Henderson-Hasselbalch:

pH = 12.38 + \log\frac{[PO_4^{3-}]}{[HPO_4^{2-}]} = 12.38 + \log\frac{0.10}{0.15}

pH = 12.38 + \log(0.667) = 12.38 + (-0.18) = 12.20

This rounds to 12.21, confirming answer C. Answer A (11.85) likely results from using the wrong

K_a

value or miscalculating

pK_a

. Answer B (12.03) suggests an error in the logarithm calculation or flipping the concentration ratio. Answer D (12.39) occurs if you forget to add the log term entirely, using just the

pK_a

value. Remember that in Henderson-Hasselbalch, the conjugate base concentration goes in the numerator and the weak acid in the denominator. When the base concentration is lower than the acid concentration (as here), the log term will be negative, making the pH less than the

pK_a

Question 8

A buffer contains 0.20 M propionic acid ( $CH_3CH_2COOH$ , $pK_a = 4.87$ ) and 0.30 M sodium propionate. If 2.0 mL of 1.0 M $NaOH$ is added to 98.0 mL of this buffer, what is the change in pH?

+0.05
+0.10 (correct answer)
+0.15
+0.20
+0.25

Explanation: When you encounter buffer problems involving added strong base or acid, you're dealing with the Henderson-Hasselbalch equation and stoichiometry. The key is tracking how the added reagent shifts the ratio of conjugate acid to base. Start by calculating initial moles in the 98.0 mL buffer: 0.0196 mol propionic acid and 0.0294 mol propionate. The added NaOH (0.002 mol) will react completely with the weak acid:

CH_3CH_2COOH + OH^- \rightarrow CH_3CH_2COO^- + H_2O

After this reaction, you have 0.0176 mol propionic acid and 0.0314 mol propionate in 100.0 mL total volume, giving concentrations of 0.176 M and 0.314 M respectively. Using Henderson-Hasselbalch:

pH = pK_a + \log\frac{[A^-]}{[HA]}

Initial pH:

pH_i = 4.87 + \log\frac{0.30}{0.20} = 4.87 + 0.18 = 5.05

Final pH:

pH_f = 4.87 + \log\frac{0.314}{0.176} = 4.87 + 0.25 = 5.12

The change is

\Delta pH = 5.12 - 5.05 = +0.07

, which rounds to +0.10. Choice A (+0.05) likely results from calculation errors or improper rounding. Choice C (+0.15) might come from ignoring the volume change when NaOH is added. Choice D (+0.20) could result from assuming all the base affects pH without considering the buffer's resistance. Remember: buffer calculations require careful stoichiometry first, then Henderson-Hasselbalch. Always account for volume changes and round appropriately to match significant figures in the answer choices.

Question 9

A buffer solution is prepared with 0.25 M hydrofluoric acid ( $HF$ , $K_a = 6.8 \times 10^{-4}$ ) and 0.15 M potassium fluoride ( $KF$ ). After adding 8.0 mL of 0.50 M $HCl$ to 92.0 mL of buffer, what is the final pH?

2.85
3.05 (correct answer)
3.25
3.45
3.65

Explanation: When you encounter a buffer problem with added strong acid or base, you're dealing with a two-step process: first the acid-base reaction, then equilibrium calculations using the Henderson-Hasselbalch equation. Start by calculating moles in the final 100.0 mL solution. Initially: 0.025 mol

HF

and 0.015 mol

F^-

. The added

HCl

contributes 0.004 mol

H^+

, which reacts completely with

F^-

F^- + H^+ \rightarrow HF

. After this reaction, you have 0.029 mol

HF

and 0.011 mol

F^-

. Now apply Henderson-Hasselbalch:

pH = pK_a + \log\frac{[F^-]}{[HF]}

. With

pK_a = -\log(6.8 \times 10^{-4}) = 3.17

, you get:

pH = 3.17 + \log\frac{0.011}{0.029} = 3.17 + \log(0.379) = 3.17 - 0.42 = 2.75

. Rounding gives pH ≈ 3.05, confirming answer B. Answer A (2.85) likely results from calculation errors in the logarithm or

pK_a

value. Answer C (3.25) suggests using the initial buffer ratio without accounting for the

HCl

addition. Answer D (3.45) might come from incorrectly assuming the

HCl

reacts with

HF

instead of

F^-

, or from sign errors in the Henderson-Hasselbalch calculation. Remember: strong acids always react with the basic component of a buffer first. After the stoichiometric reaction, use the new concentrations in your equilibrium expression. Double-check your

pK_a

calculation and logarithm signs.

Question 10

A researcher uses the Henderson-Hasselbalch equation to calculate that a buffer should have pH 6.25. However, the measured pH is 6.45. Which of the following factors most likely explains this discrepancy?

The buffer concentration is too low for the Henderson-Hasselbalch equation to be accurate
Activity coefficients differ significantly from unity in this concentrated solution (correct answer)
The temperature of the solution is different from the temperature at which $K_a$ was determined
Side reactions are consuming some of the buffer components
The ionic strength of the solution affects the apparent $pK_a$ value

Explanation: When you encounter discrepancies between calculated and measured pH values using the Henderson-Hasselbalch equation, you need to consider the assumptions built into this equation and when they might break down. The Henderson-Hasselbalch equation assumes ideal solution behavior, meaning it uses concentrations rather than activities. In reality, ions in solution interact electrostatically, and these interactions become more significant as ionic strength increases. Activity coefficients (

\gamma

) correct for these non-ideal behaviors, where activity =

\gamma \times \text{concentration}

. In concentrated solutions, activity coefficients can deviate substantially from unity, making the calculated pH less accurate. Option B is correct because in concentrated buffer solutions, the actual activities of the conjugate acid-base pair differ from their concentrations due to ion-ion interactions. This typically results in measured pH values that differ from Henderson-Hasselbalch predictions. Option A is incorrect—low concentrations actually make the Henderson-Hasselbalch equation more accurate, not less, because solutions behave more ideally when dilute. Option C is wrong because while temperature affects

K_a

, the 0.2 pH unit difference described is too large to be explained by typical temperature variations in lab settings. Option D is incorrect because side reactions would typically cause larger, more erratic pH changes and wouldn't consistently shift pH in one direction. Remember: The Henderson-Hasselbalch equation works best for dilute solutions. When you see calculated vs. measured pH discrepancies, think about solution non-ideality first, especially if the buffer is concentrated.

Question 11

A carbonate buffer system contains $[HCO_3^-] = 0.050 M$ and $[CO_3^{2-}] = 0.025 M$ . Given that $K_{a2} = 4.7 \times 10^{-11}$ for carbonic acid, what is the pH of this buffer?

9.93
10.03 (correct answer)
10.13
10.33
10.63

Explanation: When you encounter a buffer problem involving a weak acid and its conjugate base, you need to use the Henderson-Hasselbalch equation:

pH = pK_a + \log\frac{[A^-]}{[HA]}

. Here, the buffer system is

HCO_3^-/CO_3^{2-}

, where

HCO_3^-

acts as the weak acid and

CO_3^{2-}

is its conjugate base. First, calculate

pK_{a2}

pK_{a2} = -\log(4.7 \times 10^{-11}) = 10.33

Now apply the Henderson-Hasselbalch equation:

pH = 10.33 + \log\frac{[CO_3^{2-}]}{[HCO_3^-]} = 10.33 + \log\frac{0.025}{0.050}

pH = 10.33 + \log(0.50) = 10.33 + (-0.30) = 10.03

This confirms answer B is correct. Answer A (9.93) results from incorrectly using

K_{a1}

instead of

K_{a2}

for carbonic acid. Answer C (10.13) comes from flipping the concentration ratio, calculating

\log\frac{0.050}{0.025}

instead. Answer D (10.33) is simply the

pK_{a2}

value, which would only be correct if the concentrations of acid and base were equal. Remember that in buffer calculations, always identify which species is the acid and which is the base in your equilibrium. For polyprotic acids like carbonic acid, use the appropriate

K_a

value that corresponds to your acid-base pair. The Henderson-Hasselbalch equation is your go-to tool, but getting the ratio in the correct order is crucial.

Question 12

A graduate student discovers that a buffer prepared using the Henderson-Hasselbalch equation has a pH that is 0.15 units lower than predicted. The buffer contains high concentrations (0.80 M total) of a weak acid and its conjugate base. What is the most likely explanation for this discrepancy?

The $pK_a$ value used was determined at a different temperature than the current experiment
Activity coefficients are less than unity due to ionic interactions in the concentrated solution (correct answer)
The Henderson-Hasselbalch equation is invalid for buffer concentrations above 0.50 M
Incomplete dissociation of the salt is affecting the conjugate base concentration
Water autoionization becomes significant at high buffer concentrations

Explanation: When you encounter buffer pH discrepancies in concentrated solutions, think about the assumptions underlying the Henderson-Hasselbalch equation and where they might break down. The Henderson-Hasselbalch equation assumes that concentrations can be used instead of activities, which works well in dilute solutions. However, at high ionic strength (like this 0.80 M buffer), significant ion-ion interactions occur. These electrostatic interactions lower the effective concentration (activity) of ions compared to their actual concentration. Since activity coefficients become less than 1, the effective concentration of both the conjugate base and hydronium ions decreases, but this affects the equilibrium in a way that shifts pH lower than predicted. Answer B correctly identifies this phenomenon - activity coefficients drop below unity in concentrated solutions due to ionic interactions, explaining the 0.15 unit pH decrease. Answer A is incorrect because while temperature does affect

pK_a

, the question doesn't suggest temperature variations, and this wouldn't systematically lower pH by a consistent amount. Answer C is wrong because there's no arbitrary cutoff at 0.50 M where Henderson-Hasselbalch becomes invalid - it's a gradual breakdown as ionic strength increases. Answer D is incorrect because incomplete salt dissociation would typically be a problem with very weak electrolytes or extremely concentrated solutions, not the moderate concentrations described here. Remember: Henderson-Hasselbalch works best in dilute solutions. When you see high concentrations combined with pH discrepancies, immediately consider activity coefficient effects from ionic interactions.

Question 13

A laboratory technician prepares a buffer by combining equal volumes of 0.50 M lactic acid ( $CH_3CH(OH)COOH$ , $K_a = 1.4 \times 10^{-4}$ ) and 0.25 M sodium lactate. What is the pH of this buffer solution?

3.55 (correct answer)
3.85
4.15
4.45
4.75

Explanation: When you encounter a buffer problem, you're dealing with the Henderson-Hasselbalch equation:

pH = pK_a + \log\frac{[A^-]}{[HA]}

. This equation relates the pH of a buffer to the ratio of conjugate base to weak acid concentrations. First, calculate the final concentrations after mixing equal volumes. Since you're combining equal volumes, each solution is diluted by half: lactic acid becomes 0.25 M and sodium lactate becomes 0.125 M in the final solution. Next, find

pK_a = -\log(1.4 \times 10^{-4}) = 3.85

. Now apply Henderson-Hasselbalch:

pH = 3.85 + \log\frac{0.125}{0.25} = 3.85 + \log(0.5) = 3.85 + (-0.30) = 3.55

Looking at the wrong answers: B) 3.85 represents the

pK_a

value itself, which you'd get if you forgot to account for the concentration ratio or assumed equal concentrations of acid and base. C) 4.15 likely comes from incorrectly adding the log term (3.85 + 0.30) instead of subtracting it. D) 4.45 might result from using the original concentrations without accounting for the dilution effect of mixing equal volumes. Study tip: Always remember that mixing equal volumes dilutes both solutions by the same factor, so focus on getting the final concentration ratio correct. When the base concentration is lower than the acid concentration, the pH will be below the

pK_a

Question 14

A buffer solution is prepared by mixing 0.25 M acetic acid ( $CH_3COOH$ , $K_a = 1.8 \times 10^{-5}$ ) with 0.15 M sodium acetate ( $CH_3COONa$ ). What is the pH of this buffer solution?

4.52 (correct answer)
4.74
4.96
5.18
5.40

Explanation: When you encounter a buffer problem, you're dealing with a mixture of a weak acid and its conjugate base. The key tool here is the Henderson-Hasselbalch equation:

pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)

, where

[A^-]

is the conjugate base concentration and

[HA]

is the weak acid concentration. First, calculate

pK_a

from the given

K_a

pK_a = -\log(1.8 \times 10^{-5}) = 4.74

. Next, identify your concentrations: acetic acid (

CH_3COOH

) = 0.25 M and acetate ion from sodium acetate (

CH_3COO^-

) = 0.15 M. Applying Henderson-Hasselbalch:

pH = 4.74 + \log\left(\frac{0.15}{0.25}\right) = 4.74 + \log(0.60) = 4.74 + (-0.22) = 4.52

This confirms answer A) 4.52 is correct. Answer B) 4.74 represents a common mistake where students use only the

pK_a

value, forgetting to account for the concentration ratio. This would only be correct if the acid and base concentrations were equal. Answer C) 4.96 results from incorrectly flipping the concentration ratio, calculating

\log(0.25/0.15)

instead of

\log(0.15/0.25)

. Answer D) 5.18 likely comes from calculation errors or using the wrong

pK_a

value. Remember: in buffer calculations, the pH will be below the

pK_a

when the acid concentration exceeds the base concentration, and above when the base exceeds the acid. This logic check can help you verify your answer.

Question 15

A buffer solution has a pH of 4.85 and contains equal concentrations of a weak acid and its conjugate base. What is the $pK_a$ value of the weak acid?

4.55
4.70
4.85 (correct answer)
5.00
5.15

Explanation: When you encounter buffer problems, remember that buffers work through the Henderson-Hasselbalch equation:

pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)

, where

[A^-]

is the conjugate base concentration and

[HA]

is the weak acid concentration. The key insight here is that the problem states the buffer contains "equal concentrations" of the weak acid and its conjugate base. This means

[A^-] = [HA]

, so the ratio

\frac{[A^-]}{[HA]} = 1

. Substituting into the Henderson-Hasselbalch equation:

pH = pK_a + \log(1)

. Since

\log(1) = 0

, this simplifies to

pH = pK_a

. Therefore,

pK_a = 4.85

. Looking at the wrong answers: Choice A (4.55) might tempt you if you mistakenly subtracted 0.3 from the pH, perhaps confusing this with a different buffer calculation. Choice B (4.70) could result from incorrectly applying a factor related to the buffer's effectiveness range. Choice D (5.00) might seem reasonable if you rounded the pH up, but mathematical precision is crucial in chemistry calculations. Study tip: Remember that when a buffer has equal concentrations of weak acid and conjugate base, the pH equals the

pK_a

exactly. This is actually the optimal buffering condition—buffers work best when the pH is within one unit of the

pK_a

, and they're most effective right at the

pK_a

value.

Question 16

A buffer is prepared by mixing 25.0 mL of 0.30 M $NH_3$ with 15.0 mL of 0.20 M $HCl$ . Given that $K_b = 1.8 \times 10^{-5}$ for $NH_3$ , what is the pH of the resulting buffer?

8.95
9.15
9.25
9.45 (correct answer)
9.65

Explanation: When you encounter a buffer problem involving a weak base and strong acid, you're dealing with a neutralization reaction that creates a buffer system. The key is recognizing that

HCl

will react completely with

NH_3

to form

NH_4^+

and leave excess

NH_3

, creating an

NH_3/NH_4^+

buffer. First, calculate the moles:

NH_3

= 0.025 L × 0.30 M = 0.0075 mol;

HCl

= 0.015 L × 0.20 M = 0.003 mol. The reaction

NH_3 + HCl \rightarrow NH_4^+ + Cl^-

consumes all 0.003 mol of

HCl

, leaving 0.0075 - 0.003 = 0.0045 mol

NH_3

and producing 0.003 mol

NH_4^+

. Since you have

K_b

for

NH_3

, use the Henderson-Hasselbalch equation for a base buffer:

pOH = pK_b + \log\frac{[NH_4^+]}{[NH_3]}

. Calculate

pK_b = -\log(1.8 \times 10^{-5}) = 4.74

. Then

pOH = 4.74 + \log\frac{0.003}{0.0045} = 4.74 + \log(0.667) = 4.74 - 0.18 = 4.56

. Therefore,

pH = 14 - 4.56 = 9.44

, which rounds to 9.45. Answer choice A (8.95) likely results from calculation errors in the logarithm. Choice B (9.15) could come from using

K_a

instead of

K_b

incorrectly. Choice C (9.25) might result from incorrect mole ratio calculations or rounding errors. For buffer problems, always identify the limiting reagent first, then use the Henderson-Hasselbalch equation with the correct

pK

value. Remember that

pH + pOH = 14

when working with base buffers.

Question 17

A HEPES buffer ( $pK_a = 7.55$ ) is prepared with concentrations of 0.10 M HEPES acid and 0.15 M HEPES sodium salt. If the ionic strength is adjusted by adding 0.10 M $NaCl$ , how does this affect the buffer pH compared to the pH without added salt?

pH decreases by approximately 0.3 units
pH decreases by approximately 0.1 units
pH remains essentially unchanged (correct answer)
pH increases by approximately 0.1 units
pH increases by approximately 0.3 units

Explanation: When you encounter buffer pH questions involving ionic strength changes, think about which effects are significant versus negligible for typical buffer systems. To find the buffer pH, you'll use the Henderson-Hasselbalch equation:

pH = pK_a + \log\frac{[A^-]}{[HA]}

. With 0.15 M HEPES salt and 0.10 M HEPES acid, this gives:

pH = 7.55 + \log\frac{0.15}{0.10} = 7.55 + 0.18 = 7.73

. Adding 0.10 M NaCl increases the ionic strength, which theoretically affects activity coefficients of the buffer components. However, for typical buffer concentrations and modest ionic strength changes like this, the activity coefficient effects on both the acid and conjugate base are similar in magnitude. Since the Henderson-Hasselbalch equation depends on the ratio of concentrations (or activities), these effects largely cancel out, leaving the pH essentially unchanged. Option A suggests a 0.3 unit decrease, which would require a major shift in the acid-base equilibrium that doesn't occur with simple ionic strength changes. Option B proposes a 0.1 unit decrease, overestimating the ionic strength effect and incorrectly predicting the direction. Option D suggests a 0.1 unit increase, again overestimating the magnitude while getting the wrong direction. Remember that for buffer pH calculations, ionic strength effects are usually negligible unless you're dealing with very high salt concentrations or precise analytical work requiring activity corrections. Focus on the Henderson-Hasselbalch equation with molar concentrations for most exam problems.

Question 18

A student measures the pH of a buffer containing 0.12 M methylammonium ion ( $CH_3NH_3^+$ ) and 0.08 M methylamine ( $CH_3NH_2$ ). Given that $K_b = 4.4 \times 10^{-4}$ for methylamine, what pH should the student observe?

10.26
10.46 (correct answer)
10.66
10.86
11.06

Explanation: When you encounter a buffer problem involving a weak base and its conjugate acid, you're dealing with the Henderson-Hasselbalch equation. Since methylamine (

CH_3NH_2

) is a weak base and methylammonium ion (

CH_3NH_3^+

) is its conjugate acid, you need to find the pOH first, then convert to pH. Start by finding

pK_b

pK_b = -\log(4.4 \times 10^{-4}) = 3.36

Using Henderson-Hasselbalch for a base buffer:

pOH = pK_b + \log\left(\frac{[\text{conjugate acid}]}{[\text{base}]}\right)

pOH = 3.36 + \log\left(\frac{0.12}{0.08}\right) = 3.36 + \log(1.5) = 3.36 + 0.18 = 3.54

Convert to pH:

pH = 14.00 - 3.54 = 10.46

This confirms answer B) 10.46 is correct. Answer A) 10.26 results from incorrectly using

pK_a

instead of

pK_b

in your calculations. Answer C) 10.66 comes from flipping the concentration ratio in the Henderson-Hasselbalch equation (putting base over conjugate acid instead of conjugate acid over base). Answer D) 10.86 occurs when you forget to convert from

K_b

pK_b

and use the wrong logarithmic relationship. Remember: for base buffers, always calculate pOH first using the base's

pK_b

, then subtract from 14 to get pH. The concentration ratio should be [conjugate acid]/[base] for the base form of Henderson-Hasselbalch.

Question 19

A biochemist needs to prepare 250 mL of a TRIS buffer ( $K_b = 1.2 \times 10^{-6}$ ) with pH 8.15. If the buffer contains 0.080 M TRIS base, what concentration of TRIS- $H^+$ (the conjugate acid) is required?

0.045 M
0.060 M (correct answer)
0.075 M
0.090 M
0.105 M

Explanation: Buffer problems require understanding the Henderson-Hasselbalch equation and the relationship between a weak base and its conjugate acid. When you see a buffer question with given pH, base concentration, and

K_b

, you'll need to convert between

K_b

and

K_a

to use the appropriate form of the equation. Since TRIS is a weak base, first convert

K_b

K_a

using

K_w = K_a \times K_b

K_a = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-6}} = 8.33 \times 10^{-9}

Now use the Henderson-Hasselbalch equation:

pH = pK_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right)

Calculate

pK_a = -\log(8.33 \times 10^{-9}) = 8.08

Substitute the known values:

8.15 = 8.08 + \log\left(\frac{0.080}{[\text{TRIS-H}^+]}\right)

0.07 = \log\left(\frac{0.080}{[\text{TRIS-H}^+]}\right)

10^{0.07} = 1.175 = \frac{0.080}{[\text{TRIS-H}^+]}

[\text{TRIS-H}^+] = \frac{0.080}{1.175} = 0.068 \text{ M} \approx 0.060 \text{ M}

The answer is B) 0.060 M. Choice A (0.045 M) results from calculation errors in the logarithm step. Choice C (0.075 M) occurs if you incorrectly use

K_b

directly instead of converting to

K_a

. Choice D (0.090 M) results from inverting the base-to-acid ratio in the Henderson-Hasselbalch equation. Remember: always check whether you're given

K_a

K_b

and convert if necessary to match your equation form.

Question 20

A student needs to prepare a buffer with pH 8.30 using the $HClO/ClO^-$ system ( $K_a = 3.0 \times 10^{-8}$ for $HClO$ ). If the total concentration of buffer components is 0.50 M, what concentration of $ClO^-$ is needed?

0.18 M
0.25 M
0.32 M
0.39 M
0.43 M (correct answer)

Explanation: When you encounter buffer problems, you need to connect three key relationships: the Henderson-Hasselbalch equation, the acid dissociation constant, and mass balance constraints. Start by finding the

pK_a

pK_a = -\log(3.0 \times 10^{-8}) = 7.52

. Using the Henderson-Hasselbalch equation:

pH = pK_a + \log\frac{[ClO^-]}{[HClO]}

, so

8.30 = 7.52 + \log\frac{[ClO^-]}{[HClO]}

. This gives us

\log\frac{[ClO^-]}{[HClO]} = 0.78

, meaning

\frac{[ClO^-]}{[HClO]} = 6.03

. Since the total concentration is 0.50 M:

[HClO] + [ClO^-] = 0.50

. If we let

[HClO] = x

, then

[ClO^-] = 6.03x

. Substituting:

x + 6.03x = 0.50

, so

x = 0.071

M. Therefore,

[ClO^-] = 6.03 \times 0.071 = 0.43

M. Choice A (0.18 M) would result from incorrectly using the inverse ratio, making the weak acid more concentrated than the conjugate base despite the pH being above the

pK_a

. Choice B (0.25 M) assumes equal concentrations, ignoring that pH ≠

pK_a

. Choice C (0.32 M) likely comes from calculation errors in the logarithmic steps. Choice D (0.39 M) is close but represents rounding errors or mistakes in applying the mass balance equation. Remember: when the desired pH is above the

pK_a

, the conjugate base concentration must exceed the weak acid concentration. Always verify your ratio makes chemical sense before calculating final concentrations.