Henderson–Hasselbalch Equation - College Chemistry

1. 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?
   Answer: 3.31 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 $[H_3Cit] = 0.08$ M: $pH = 3.13 + \log\frac{0.12}{0.08} = 3.13 + \log(1.5) = 3.13 + 0.18 = 3.31$ 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.
2. 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?
   Answer: 0.060 M 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 $K_b$ to $K_a$ 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 $K_a$ or $K_b$ and convert if necessary to match your equation form.
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?
   Answer: 4.02 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.
4. 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?
   Answer: 4.02 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.
5. 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?
   Answer: 9.45 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.
6. 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?
   Answer: 4.85 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.
7. 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?
   Answer: 3.05 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.
8. 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?
   Answer: pH remains essentially unchanged 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.
9. 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?
   Answer: 10.46 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 $K_b$ to $pK_b$ 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.
10. 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?
    Answer: +0.10 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.
11. 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?
    Answer: 4.50 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.
12. 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?
    Answer: 12.21 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 $pK_a$.
13. 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?
    Answer: 10.03 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.
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?
    Answer: 4.52 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.
15. 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?
    Answer: Activity coefficients differ significantly from unity in this concentrated solution 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.
16. 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?
    Answer: 3.55 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 $pK_a$.
17. 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 $\frac{[NH_2CH_2COO^-]}{[^+NH_3CH_2COO^-]}$?
    Answer: 3.2 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.
18. An imidazole buffer system has $pK_a = 7.00$. To achieve a buffer capacity that is 80% of the maximum possible, what should be the ratio $\frac{[Im]}{[ImH^+]}$ where Im is imidazole and $ImH^+$ is the protonated form?
    Answer: 1.00 Buffer capacity questions test your understanding of the Henderson-Hasselbalch equation and how buffer effectiveness relates to the ratio of conjugate acid-base pairs. Buffer capacity is maximized when the concentrations of the weak acid and its conjugate base are equal, which occurs at pH = pKₐ. At this point, the system can most effectively resist pH changes in either direction. However, buffers maintain good capacity within about 1 pH unit of the pKₐ value. For 80% maximum buffer capacity, you need to find where the buffer maintains this effectiveness. This occurs when the pH differs from pKₐ by approximately 0.3 units (either 0.3 above or below). Using the Henderson-Hasselbalch equation: $pH = pK_a + \log\frac{[Im]}{[ImH^+]}$ For optimal capacity at pKₐ = 7.00, when pH = pKₐ: $7.00 = 7.00 + \log\frac{[Im]}{[ImH^+]}$, so $\log\frac{[Im]}{[ImH^+]} = 0$, giving us $\frac{[Im]}{[ImH^+]} = 1.00$. Choice A (0.25) represents a system heavily favoring the protonated form, operating at pH ≈ 6.4, where buffer capacity is significantly reduced. Choice B (0.50) corresponds to pH ≈ 6.7, still below optimal capacity. Choice D (2.00) represents pH ≈ 7.3, also away from maximum effectiveness. Remember: Maximum buffer capacity always occurs when the concentrations of conjugate acid-base pairs are equal (ratio = 1), regardless of the specific pKₐ value. This is a fundamental principle worth memorizing.
19. 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?
    Answer: 7.38 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.
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?
    Answer: 0.43 M 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.

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