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?单项选择题

类似问题

Write an expression for the acid ionization constant (Ka) for the dissociation of the following weak acid H2CO3.

A 0.10 M solution of a weak acid, HX, is 0.059% ionized. Evaluate Ka for the acid.

The pH of a solution of a 0.15 M solution of HOCl is 4.14. What is the Ka for HOCl?

For the dissociation of a very weak acid, HA, in water

The larger the value of Ka

A 0.025 M solution of a weak acid has a percent ionization of 3.2%. What is the Ka of the weak acid?

Based on the following titration curve, what is the Ka for this weak acid?

Which of the following trees corresponds to a potential parse of the ambiguous sentence below, with correct syntactic categories? Some diagnostics are provided.

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单