Henderson-Hasselbalch Equation
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AP Chemistry › Henderson-Hasselbalch Equation
A buffer is made from benzoic acid and benzoate, $\text{C}_6\text{H}_5\text{COOH}/\text{C}_6\text{H}_5\text{COO}^-$. If $\text{C}_6\text{H}_5\text{COO}^-=0.040\ \text{M}$ and $\text{C}_6\text{H}_5\text{COOH}=0.010\ \text{M}$ and $pK_a=4.20$, what is the pH?
3.60
4.20
4.60
4.80
5.20
Explanation
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.
Two buffers are prepared using the same conjugate acid-base pair $\text{HA}/\text{A}^-$. For Buffer 1, $\text{A}^-/\text{HA}=10$. For Buffer 2, $\text{A}^-/\text{HA}=0.10$. Which statement correctly compares the pH values?
pH(Buffer 1) is 1 unit higher than pH(Buffer 2)
pH(Buffer 1) is 2 units lower than pH(Buffer 2)
pH(Buffer 1) is 2 units higher than pH(Buffer 2)
pH(Buffer 1) is 1 unit lower than pH(Buffer 2)
pH(Buffer 1) equals pH(Buffer 2)
Explanation
This question tests understanding of how concentration ratios affect pH differences in buffers. Using the Henderson-Hasselbalch equation, Buffer 1 has pH = pKa + log(10) = pKa + 1, while Buffer 2 has pH = pKa + log(0.10) = pKa + (-1) = pKa - 1. The difference is (pKa + 1) - (pKa - 1) = 2 pH units, with Buffer 1 being higher. Choice B is incorrect because it underestimates the pH difference by not recognizing that log(10) = 1 and log(0.10) = -1. When comparing buffers with reciprocal concentration ratios, the pH difference equals 2 times the log of the ratio.
A buffer is made using the acid/base pair $\text{HCO}_3^- / \text{CO}_3^{2-}$. If $\text{HCO}_3^-=0.50\ \text{M}$ and $\text{CO}_3^{2-}=0.50\ \text{M}$ and $pK_a(\text{HCO}_3^-)=10.33$, what is the pH of the buffer?
9.33
9.83
10.33
10.83
11.33
Explanation
This question tests the Henderson-Hasselbalch equation when acid and base concentrations are equal. In the HCO₃⁻/CO₃²⁻ system, HCO₃⁻ acts as the acid and CO₃²⁻ as the base, so pH = pKa + log([CO₃²⁻]/[HCO₃⁻]). With equal concentrations (0.50 M each), pH = 10.33 + log(0.50/0.50) = 10.33 + log(1) = 10.33 + 0 = 10.33. Choice C (10.83) is incorrect because it adds 0.5 instead of log(1) = 0 to the pKa. When concentrations are equal in a buffer, the pH equals the pKa because log(1) = 0.
A buffer solution contains $\text{HClO}$ and $\text{ClO}^-$. If $\text{HClO}=0.30\ \text{M}$ and $\text{ClO}^-=0.10\ \text{M}$ and $pK_a(\text{HClO})=7.53$, what is the pH of the buffer?
6.53
7.05
7.53
7.83
8.01
Explanation
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).
A buffer contains $\text{NH}_3$ and $\text{NH}_4^+$. The solution has $\text{NH}_3=0.30\ \text{M}$ and $\text{NH}_4^+=0.10\ \text{M}$. Given $pK_a(\text{NH}_4^+)=9.25$, what is the pH of the buffer?
8.25
8.77
9.25
9.73
10.23
Explanation
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.
A buffer contains $0.10\ \text{M}$ $\text{H}_2\text{PO}_4^-$ and $0.20\ \text{M}$ $\text{HPO}_4^{2-}$. Given $\text{p}K_a(\text{H}_2\text{PO}_4^-)=7.21$, what is the pH of the buffer?
6.71
6.91
7.21
7.51
14.42
Explanation
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.
Two buffers use the same conjugate pair, $\text{HF}/\text{F}^-$. Buffer X has $\text{HF}=0.30\ \text{M}$ and $\text{F}^-=0.10\ \text{M}$. Buffer Y has $\text{HF}=0.30\ \text{M}$ and $\text{F}^-=0.30\ \text{M}$. Given $\text{p}K_a(\text{HF})=3.17$, which statement is correct?
pH(Buffer X) < pH(Buffer Y)
pH(Buffer Y) = 2,$\text{p}$K_a
pH(Buffer X) = pH(Buffer Y)
pH(Buffer X) > pH(Buffer Y)
pH(Buffer X) = \t$\frac{1}{2}$,$\text{p}$K_a
Explanation
This question tests the application of the Henderson-Hasselbalch equation to compare pH of buffers with varying ratios. For Buffer X, pH = 3.17 + log(0.10/0.30) ≈ 3.17 - 0.48 = 2.69; for Y, pH = 3.17 + log(1) = 3.17. Thus, pH(X) < pH(Y) due to lower ratio in X. Equal acid but differing base affects pH. A tempting distractor is equal pH, from misconception that same [HF] means same pH. Focus on the ratio in the log term of the Henderson-Hasselbalch equation for comparisons.
A buffer contains $0.20\ \text{M}$ $\text{HSO}_4^-$ and $0.20\ \text{M}$ $\text{SO}_4^{2-}$. Given $\text{p}K_a(\text{HSO}_4^-)=1.99$, what is the pH of the buffer?
0.00
1.69
1.99
2.29
3.98
Explanation
This question tests the application of the Henderson-Hasselbalch equation to an equimolar buffer in a strong acid system. The buffer has HSO4- (HA) and $SO4^2$- (A-) both at 0.20 M, with pKa = 1.99. pH = 1.99 + log(1) = 1.99. Equal concentrations make pH equal to pKa. A tempting distractor is 3.98, possibly from doubling pKa misconception. Remember, the Henderson-Hasselbalch equation simplifies to pKa when [A-] = [HA], a useful check for balanced buffers.
A buffer contains $0.25\ \text{M}$ lactic acid, $\text{HLac}$, and $0.50\ \text{M}$ lactate ion, $\text{Lac}^-$. Given $\text{p}K_a(\text{HLac})=3.86$, what is the pH of the buffer?
3.56
3.86
4.16
4.46
7.72
Explanation
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.
A buffer is prepared by mixing $0.20\ \text{M}$ acetic acid, $\text{HC}_2\text{H}_3\text{O}_2$, and $0.10\ \text{M}$ sodium acetate, $\text{NaC}_2\text{H}_3\text{O}_2$. Given $\text{p}K_a(\text{HC}_2\text{H}_3\text{O}_2)=4.76$, what is the pH of the buffer?
4.16
4.46
4.76
5.06
9.52
Explanation
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.