Calculate the percent ionization of a 0.125 M aqueous HCN solution. For HCN, Ka = 4.9
×
10–10.

Question

Calculate the percent ionization of a 0.125 M aqueous HCN solution.  For HCN, Ka = 4.9 
×
 10–10.

BlackTom AI · Accepted Answer

We start by identifying the key values: the problem gives a weak acid HCN with Ka = 4.9 × 10^(-10) and initial concentration C0 = 0.125 M. The percent ionization is defined as ([H3O+] / C0) × 100, since for a monoprotic acid, [H+] = [A-].

Option analysis begins with the general approach for a weak acid: for small degrees of ionization, Ka ≈ x^2 / (C0 − x), where x = [H+]. If x is much smaller than C0, we can approximate Ka ≈ x^2 / C0, giving x ≈ sqrt(Ka × C0).

Compute x: x ≈ sqrt((4.9 × 10^(-10)) × 0.125).
- Multiply Ka by C0: 4.9 × 10^(-10) × 0.125 = 6.125 × 10^(-11).
- Take the square root: sqrt(6.125 × 10^(-11)) ≈ sqrt(6.125) × sqrt(10^(-11)) ≈ 2.475 × 10^(-5.5).
- Evaluating 10^(-5.5) = 10^(-5) × 10^(-0.5) ≈ 1.0×10^(-5) × 0.3162 ≈ 3.162 × 10^(-6). Multiplying by 2.475 gives x ≈ 7.8 × 10^(-6) M.

Now determine percent ionization: (x / C0) × 100 = (7.8 × 10^(-6) / 0.125) × 100.
- Divide: 7.8 × 10^(-6) / 0.125 = 6.24 × 10^(-5).
- Multiply by 100: 6.24 × 10^(-5) × 100 = 6.24 × 10^(-3) = 0.00624%.

With this result in mind, examine each option:

- Option A: 0.018% — This is larger than the calculated value by about a factor of 3. The discrepancy likely arises from treating the problem as if a larger fraction ionizes, or from not using the approximation x ≪ C0. It indicates an overestimation of [H+] relative to the initial concentration.

- Option B: 0.35% — This is an even bigger departure. It would imply x/C0 ≈ 0.0035, i.e., [H+] ≈ 0.0004375 M, which would require Ka to be far larger or C0 to be much smaller. Neither is consistent with the given Ka and concentration.

- Option C: 0.51% — Similar issue as B, representing a substantially higher degree of ionization than expected for a very weak acid with Ka on the order of 10^(-10) and a modest 0.125 M concentration. The math shows [H+] would be around 10^(-3) M, which is not supported by the Ka value.

- Option D: 0.0063% — This is the only value that matches the calculation outcome from x ≈ sqrt(Ka × C0) and the percent ionization formula. The small degree of ionization is consistent with a weak acid with a very small Ka, and the arithmetic yields a result in this neighborhood.

In summary, the method relies on recognizing that HCN is a very weak acid, so its ionization is minimal and can be approximated with x ≈ sqrt(Ka × C0). Each incorrect option reflects either overestimation of ionization (A, B, C) due to not applying the weak-acid approximation or misinterpreting the relationship between Ka, C0, and [H+]. The calculation steps above lead to a value of roughly 7.8 × 10^(-6) M for [H+], giving a percent ionization of about 0.0062–0.0063%, which aligns with the last option.

Calculate the percent ionization of a 0.125 M aqueous HCN solution. For HCN, Ka = 4.9 × 10–10.

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