The pH of a 0.800 M aqueous solution of a weak acid is 1.99. What is the value of Ka for this acid?

Question

The pH of a 0.800 M aqueous solution of a weak acid is 1.99.  What is the value of Ka for this acid?

BlackTom AI · Accepted Answer

We start by identifying the given data and the formula we’ll use. The solution is 0.800 M of a weak acid, and the pH is 1.99, so the hydrogen ion concentration [H+] is 10^(-1.99). This value will be our x in the usual weak-acid equilibrium expression HA ⇌ H+ + A-. The equilibrium expression for Ka is Ka = [H+][A-] / [HA], but since [H+] ≈ [A-] = x and [HA] ≈ C0 − x, we get Ka ≈ x^2 / (C0 − x). Given that x is small compared to C0 (0.0102 M vs 0.800 M), we can approximate Ka ≈ x^2 / C0.

Step-by-step evaluation of each option:

Option 1: 1.2 × 10^−10
- If Ka were as small as 1.2e−10, the acid would be extremely weak and the resulting [H+] would be near zero for a 0.800 M solution, yielding a pH around 14; this is far from 1.99. This value is inconsistent with the observed pH and the magnitude of dissociation implied by the given pH.

Option 2: 3.6 × 10^−4
- A Ka of 3.6e−4 would correspond to a somewhat stronger weak acid. Using Ka ≈ x^2 / C0, x ≈ sqrt(Ka × C0) ≈ sqrt(3.6e−4 × 0.800) ≈ sqrt(2.88e−4) ≈ 0.0170 M, which would give pH ≈ −log(0.017) ≈ 1.77. That pH is not equal to the given 1.99, making this option inconsistent with the data.

Option 3: 1.3 × 10^−4
- This value yields x ≈ sqrt(Ka × C0) ≈ sqrt(1.3e−4 × 0.800) ≈ sqrt(1.04e−4) ≈ 0.0102 M. The corresponding pH would be −log(0.0102) ≈ 1.99, which matches the given pH very well. This is consistent with the approximation that x ≈ [H+] and x ≪ C0, so Ka ≈ x^2 / C0.

Option 4: 1.0 × 10^−2
- Ka of 1.0e−2 is quite large for a weak acid and would imply a much larger fraction of dissociation at 0.800 M, leading to a much higher [H+] than 0.0102 M and a far lower pH than 1.99. This is not compatible with the reported pH.

Option 5: 2.7 × 10^−5
- Ka = 2.7e−5 would give x ≈ sqrt(Ka × C0) ≈ sqrt(2.7e−5 × 0.800) ≈ sqrt(2.16e−5) ≈ 0.00465 M, yielding pH ≈ −log(0.00465) ≈ 2.33. This is noticeably different from 1.99, so this option does not fit the data.

In summary, comparing the calculated pH implications from each Ka with the given pH of 1.99 shows that only Ka ≈ 1.3 × 10^−4 produces the correct hydrogen-ion concentration and matches the provided information.

The pH of a 0.800 M aqueous solution of a weak acid is 1.99. What is the value of Ka for this acid?Single choice

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