What is the pH of a buffer that is 0.120 M in formic acid (HCO2H) and 0.080 M in potassium formate (HCO2K)? For formic acid, Ka = 1.8
×
10–4.

Question

What is the pH of a buffer that is 0.120 M in formic acid (HCO2H) and 0.080 M in potassium formate (HCO2K)? For formic acid, Ka = 1.8 
×
 10–4.

BlackTom AI · Accepted Answer

We start by identifying the buffering system: formic acid (HA) and its conjugate base formate (A−). The pH of a buffer can be estimated with Henderson–Hasselbalch: pH = pKa + log([A−]/[HA]).

First, convert Ka to pKa: Ka = 1.8 × 10^−4. pKa = −log10(1.8 × 10^−4) ≈ 3.745.

Next, compute the ratio [A−]/[HA] from the given concentrations: [A−] = 0.080 M, [HA] = 0.120 M, so [A−]/[HA] = 0.080/0.120 = 0.6667.

Find the logarithm: log10(0.6667) ≈ −0.176.

Now plug into the Henderson–Hasselbalch equation: pH ≈ 3.745 + (−0.176) = 3.569, which rounds to about 3.57.

Option-by-option analysis:
- Option 3.57: This value matches the calculation exactly when rounding pKa and the log ratio appropriately, so this aligns with the Henderson–Hasselbalch result.
- Option 2.33: This is significantly lower than the calculated pH. It would imply a much smaller ratio [A−]/[HA] or a much lower pKa, neither of which fits the given concentrations or Ka.
- Option 3.74: This is essentially the pKa value alone. It would occur if [A−] ≈ [HA], giving log([A−]/[HA]) ≈ 0, but here [A−]/[HA] = 0.6667, so the pH should be below pKa by about 0.18 units, not equal to pKa.
- Option 3.91: This is higher than the calculated pH. It would require a larger [A−]/[HA] ratio (closer to or above 1) or a higher pKa, neither of which applies to the given data.

In summary, applying the buffer equation with the provided Ka and concentrations yields a pH near 3.57; the other options diverge from this result due to incorrect assumptions about the ratio or the acid’s strength.

What is the pH of a buffer that is 0.120 M in formic acid (HCO2H) and 0.080 M in potassium formate (HCO2K)? For formic acid, Ka = 1.8 × 10–4.单项选择题

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