Calculate the percent ionization of HCO2H in an aqueous solution containing 0.13 M formic acid, HCO2H(aq), and 0.11 M potassium formate, HCO2K(aq). For formic acid, Ka = 1.8
×
10–4.

Question

Calculate the percent ionization of HCO2H in an aqueous solution containing 0.13 M formic acid, HCO2H(aq), and 0.11 M potassium formate, HCO2K(aq).  For formic acid, Ka = 1.8 
×
 10–4.

BlackTom AI · Accepted Answer

We need to evaluate the percent ionization of formic acid (HCO2H) in a buffered solution containing 0.13 M formic acid and 0.11 M potassium formate, given Ka = 1.8 × 10^-4.

First, restate the setup: initial [HA] = 0.13 M and initial [A-] (from the salt) = 0.11 M. The acid dissociation is HA ⇌ H+ + A-, with Ka = [H+][A-]/[HA]. In a buffer, the Henderson–Hasselbalch relation can be used to estimate pH: pH = pKa + log([A-]/[HA]).

Calculate pKa: pKa = -log10(1.8 × 10^-4) ≈ 3.745.
Compute the ratio [A-]/[HA] = 0.11 / 0.13 ≈ 0.846, and log(0.846) ≈ -0.072. Thus pH ≈ 3.745 + (-0.072) ≈ 3.673.

From pH, find [H+]: [H+] ≈ 10^(-3.673) ≈ 2.1 × 10^-4 M.

Next, relate [H+] to the extent of dissociation x. Let x be the amount of HA that dissociates: at equilibrium [H+] ≈ x (assuming water’s contribution is negligible), [A-] ≈ 0.11 + x, [HA] ≈ 0.13 - x.
Using Ka = [H+][A-]/[HA], we have Ka ≈ x(0.11 + x)/(0.13 - x).
Given that x is small relative to 0.13 and the computed [H+] ≈ 2.1 × 10^-4, we can approximate (0.11 + x) ≈ 0.11 and (0.13 - x) ≈ 0.13, so Ka ≈ x × 0.11 / 0.13.
Solving for x with Ka ≈ 1.8 × 10^-4 gives x ≈ (Ka × 0.13) / 0.11 ≈ (1.8 × 10^-4 × 0.13) / 0.11 ≈ 2.1 × 10^-4 M, which matches the [H+] estimate.

Percent ionization is defined as (amount dissociated / initial HA) × 100% = x / 0.13 × 100% ≈ (2.1 × 10^-4 / 0.13) × 100% ≈ 0.16%.

Now evaluate the given options:
- 2.1%: This would require a dissociation fraction about 0.021, which is far larger than the calculated x ≈ 2.1 × 10^-4 M. The buffer composition and Ka do not support such a high ionization; the Henderson–Hasselbalch estimate and subsequent Ka relation both point to a much smaller fraction.
- 3.7%: Similar to the above, this implies a dissociation about 0.037 of the initial 0.13 M, i.e., x ≈ 4.8 × 10^-3 M, which contradicts the small [H+] determined from Ka and pH; the numbers are not consistent with the equilibrium we used.
- 0.16%: This aligns with the calculation above: x ≈ 2 × 10^-4 M relative to 0.13 M gives ≈0.16%. Since the buffer with substantial A- suppresses dissociation, this small value is physically reasonable.
- 0.14%: While close to 0.16%, this would require x ≈ 1.82 × 10^-4 M; the difference is within a typical rounding range, but we can check against the precise Ka and pH numbers to decide which is most consistent. The more exact calculation yields about 0.16%, making 0.14% a plausible but less accurate alternative depending on rounding.

In summary, the reasoning using Henderson–Hasselbalch to estimate pH and then Ka to link to the dissociation extent yields a percent ionization on the order of 0.16%. The options that imply several percent ionization are inconsistent with the equilibrium behavior of a weak acid in a buffered solution, and the option around 0.16% best matches the quantitative result from the steps above.

Calculate the percent ionization of HCO2H in an aqueous solution containing 0.13 M formic acid, HCO2H(aq), and 0.11 M potassium formate, HCO2K(aq). For formic acid, Ka = 1.8 × 10–4.单项选择题

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