This question tests the Henderson-Hasselbalch equation with a 4:1 base-to-acid ratio. For the benzoic acid/benzoate buffer, pH = pKa + log([C₆H₅COO⁻]/[C₆H₅COOH]) = 4.20 + log(0.040/0.010) = 4.20 + log(4) = 4.20 + 0.60 = 4.80. Choice C (4.20) is incorrect because it only gives the pKa value, failing to account for the higher base concentration. Remember that log(4) ≈ 0.60, which significantly increases the pH above the pKa.

This question tests the Henderson-Hasselbalch equation when acid concentration exceeds base concentration. For the HClO/ClO⁻ buffer, pH = pKa + log([ClO⁻]/[HClO]) = 7.53 + log(0.10/0.30) = 7.53 + log(0.333) = 7.53 + (-0.48) = 7.05. Choice B (7.53) is incorrect because it ignores the concentration ratio, assuming equal amounts of acid and base. When acid concentration is higher than base concentration, the pH will be lower than the pKa by the absolute value of log(ratio).

This question tests the application of the Henderson-Hasselbalch equation to a polyprotic acid buffer system. The buffer contains H2PO4- (HA) at 0.10 M and $HPO4^2$- (A-) at 0.20 M, with pKa = 7.21. Calculating pH = 7.21 + log(0.20/0.10) = 7.21 + log(2) ≈ 7.21 + 0.30 = 7.51. This demonstrates how a higher base concentration shifts pH above pKa. A tempting distractor is 7.21, arising from the misconception of using pKa alone without the ratio. When dealing with buffers, consistently use the Henderson-Hasselbalch equation to account for concentration effects on pH.

This question tests the application of the Henderson-Hasselbalch equation to calculate the pH of an acidic buffer solution. The buffer contains lactic acid (HA) at 0.25 M and lactate ion (A-) at 0.50 M, with pKa = 3.86. Substituting gives pH = 3.86 + log(0.50/0.25) = 3.86 + log(2) ≈ 3.86 + 0.30 = 4.16. This shows the pH exceeds pKa when base concentration is higher. A tempting distractor is 3.86, from the misconception of equating pH to pKa without the log term. Remember to use the Henderson-Hasselbalch equation fully to predict buffer pH accurately across different systems.

This question tests the application of the Henderson-Hasselbalch equation to calculate the pH of an acidic buffer solution. The buffer consists of acetic acid (HA) at 0.20 M and acetate ion (A-) at 0.10 M, with pKa = 4.76. Using the equation pH = pKa + log([A-]/[HA]), we substitute to get pH = 4.76 + log(0.10/0.20) = 4.76 + log(0.5) ≈ 4.76 - 0.30 = 4.46. This shows that when the concentration of the acid is higher than the base, the pH is below the pKa value. A tempting distractor is 4.76, which arises from the misconception of equating pH directly to pKa without considering the ratio of concentrations. Always remember to use the Henderson-Hasselbalch equation by correctly identifying the conjugate acid-base pair and their concentrations for buffer pH calculations.

This question tests the application of the Henderson-Hasselbalch equation to compare buffers with extreme ratios. For Buffer 1, pH = 8.00 + log(10) = 8.00 + 1 = 9.00; for Buffer 2, pH = 8.00 + log(0.10) = 8.00 - 1 = 7.00. Thus, pH(1) > pH(2) due to higher ratio in 1. Ratios determine pH deviation. A tempting distractor is equal pH, from misconception that same pKa means same pH. Compare buffers by calculating individual pH using the Henderson-Hasselbalch equation and ratios.

This question tests the application of the Henderson-Hasselbalch equation to calculate the pH of an equimolar buffer solution. The buffer has formic acid (HA) and formate ion (A-) both at 0.050 M, with pKa = 3.75. Plugging into pH = pKa + log([A-]/[HA]) yields pH = 3.75 + log(1) = 3.75 + 0 = 3.75. In this case, equal concentrations make the pH equal to the pKa. A tempting distractor is 7.50, which might result from the misconception of doubling the pKa or confusing it with neutral pH. A useful strategy is to recognize that for any buffer, the pH is determined by pKa adjusted by the logarithm of the base-to-acid ratio.

This question tests the application of the Henderson-Hasselbalch equation to a phosphate buffer with low base ratio. The buffer has H2PO4- (HA) at 0.18 M and $HPO4^2$- (A-) at 0.02 M, with pKa = 7.21. pH = 7.21 + log(0.02/0.18) = 7.21 + log(1/9) ≈ 7.21 - 0.95 = 6.26. Low ratio lowers pH. A tempting distractor is 7.21, from equating to pKa misconception. Apply the Henderson-Hasselbalch equation carefully for polyprotic systems to predict pH shifts.

This question tests the application of the Henderson-Hasselbalch equation to calculate the pH of a basic buffer solution. The buffer contains ammonia (base) at 0.30 M and ammonium ion (acid) at 0.10 M, with pKa of NH4+ = 9.25. Substituting into pH = pKa + log([base]/[acid]) gives pH = 9.25 + log(0.30/0.10) = 9.25 + log(3) ≈ 9.25 + 0.48 = 9.73. This indicates that a higher base-to-acid ratio results in a pH above the pKa. A tempting distractor is 9.25, stemming from the misconception of ignoring the concentration ratio and using pKa directly as pH. To solve buffer problems effectively, always apply the Henderson-Hasselbalch equation with the appropriate form for acidic or basic buffers.

This question tests the Henderson-Hasselbalch equation for a basic buffer system. For $\text{NH}_3$/$\text{NH}_4^+$ buffers, $\text{NH}_4^+$ is the acid and $\text{NH}_3$ is the base, so $ \text{pH} = \text{pKa} + \log\left( \frac{[\text{NH}_3]}{[\text{NH}_4^+]} \right) $. Substituting the given values: $ \text{pH} = 9.25 + \log(0.30/0.10) = 9.25 + \log(3) = 9.25 + 0.48 = 9.73 $. Choice B (9.25) is incorrect because it represents the pKa value alone, failing to account for the concentration ratio. Remember that for basic buffers, the higher concentration of base relative to acid will make the pH higher than the pKa.