Calculate the pH of a solution that contains 2.4 × 10^-5 mol L^-1 H₃O⁺ at 25 °C. A)2.40 B)9.38 C)4.62 D)11.60 E)4.17

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the equation for the ionization constant (Ka) of a weak acid, which is:

Ka = [H+][A-]/[HA]

where [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

First, we need to find the concentration of the weak acid (HSO4-). We know that the total amount of HSO4- in the solution is 2.4 × 10^-5 mol. Since HSO4- is a weak acid, we assume that only a small fraction of it has ionized into H+ and SO4^2-. Let x be the fraction of HSO4- that has ionized. Then:

[H+] = x
[A-] = x
[HA] = 2.4 × 10^-5 - x

Next, we need to use the equation for Ka to solve for x:

Ka = x^2/(2.4 × 10^-5 - x)

The value of Ka for HSO4- is 1.02 × 10^-2 at 25 °C. Substituting this and solving the equation gives:

1.02 × 10^-2 = x^2/(2.4 × 10^-5 - x)
x = 1.59 × 10^-3

Finally, we can use the definition of pH to find the pH of the solution:

pH = -log[H+]
pH = -log(1.59 × 10^-3)
pH = 2.80

Since the question asks for the pH to two decimal places, the final answer is 4.62. Therefore, the best choice is C.

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