Final answer:
The calculated pH of a solution containing 0.250M HF and 0.250M HClO, focusing on HF due to its higher Ka, is approximately 2.5, which does not match any of the given options. A further clarification on the question is needed.
Step-by-step explanation:
The student has asked about the pH of a solution that contains both 0.250M HF and 0.250M HClO. HF has a higher dissociation constant (Ka = 3.05e-4) compared to HClO (Ka = 2.9e-8). Due to the much higher Ka value, the dissociation of HF will have a more significant impact on the pH than that of HClO.
Calculating the pH: Since HF is the stronger acid here (higher Ka), we will focus on it for the pH calculation. The concentration of H+ ions produced will predominantly come from HF dissociation.
The acid dissociation reaction for HF can be represented as:
HF → H+ + F-
We can write the expression for the dissociation constant (Ka) for HF as:
Ka = [H+][F-]/[HF]
However, since [HF] ≈ [H+] ≈ [F-] because of the 1:1 stoichiometry, we can simplify it to:
Ka = [H+]^2
This leads to [H+] = sqrt(Ka)
[H+] = sqrt(3.05e-4)
The pH is defined as:
pH = -log[H+]
By calculating the negative logarithm of the H+ ion concentration, we find:
pH = -log(sqrt(3.05e-4))
After the computation, the pH is approximately 2.5, which is not an option given. Since none of the given choices match our calculation and considering that HF and HClO are both weak acids and not likely to create a neutral pH, the next closest choice would be a pH of 3.0 if we consider a slight deviation due to experimental factors, or it may imply that the question has a typo. Hence, the correct pH value needs further clarification.