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

First, I restate the problem to ensure clarity: we have formic acid, HCO2H, at 0.13 M and its conjugate base formate from potassium formate at 0.11 M in solution, with Ka = 1.8 × 10^-4. The task is to find the percent ionization of formic acid in this buffered mixture.

To analyze, recognize this is a weak acid (HA) with a significant amount of its conjugate base (A−) already present. The equilibrium is HA ⇌ H+ + A−, and the acid dissociation constant is Ka = [H+][A−]/[HA]. At equilibrium, let the amount of HA that dissociates be x (in molarity). Then:
- [H+] = x
- [A−] = 0.11 + x (the 0.11 M from salt plus the dissociated amount)
- [HA] = 0.13 − x
Thus Ka = x(0.11 + x)/(0.13 − x).

Because this is a buffer with relatively large initial A−, x is small compared to the initial concentrations. A first-order approximation sets (0.11 + x) ≈ 0.11 and (0.13 − x) ≈ 0.13, giving x ≈ Ka × 0.13 / 0.11.

Compute: Ka × 0.13 / 0.11 = (1.8 × 10^-4) × (0.13 / 0.11) ≈ (1.8 × 10^-4) × 1.1818 ≈ 2.13 × 10^-4 M.

Percent ionization is defined as (amount of HA that dissociated) / (initial HA) × 100% = x / 0.13 × 100%.

Calculate: (2.13 × 10^-4) / 0.13 × 100% ≈ 0.001638 × 100%? Careful with units: 2.13 × 10^-4 divided by 0.13 equals about 1.64 × 10^-3, and multiplying by 100% gives ≈ 0.164%.

Thus, the percent ionization of formic acid under these buffered conditions is approximately 0.16%.

Notes on alternatives and potential misconceptions:
- If one ignored the 0.11 M A− from the salt and treated it as a pure solution of HA, the percent ionization would be much larger; the common ion effect dramatically suppresses ionization.
- If you tried to solve the exact equation Ka = x(0.11 + x)/(0.13 − x) without the small-x approximation, you would still obtain a value very close to 2.1 × 10^-4 for x, since x is indeed small relative to both 0.11 and 0.13, confirming the approximation’s validity.
- The presence of a buffer means pH is governed by Henderson–Hasselbalch, pH = pKa + log([A−]/[HA]); converting that to [H+] and then to percent ionization yields a consistent result near 0.16%.

Chem 1C General Chemis... Self Assessment Quiz 16

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.

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