Final answer:
The pH of a 0.75 M solution of HCN with a Ka of 6.17x10^-10 can be estimated through an equilibrium calculation, resulting in an approximate pH of 5.17, which is not listed among the provided choices.
Step-by-step explanation:
The question asks for the pH of a 0.75 M solution of HCN, given the acid dissociation constant (Ka) of HCN is 6.17 × 10-10. To solve this problem, we need to perform an equilibrium calculation. The dissociation of HCN in water can be written as:
HCN + H2O → H3O+ + CN-
We can set up the equilibrium expression as follows:
Ka = [H3O+][CN-] / [HCN]
Because HCN is a weak acid, we assume that it only partially dissociates, and if we let x be the concentration of H3O+ that forms, then:
Ka = x2 / (0.75-x)
Given that Ka is very small, the value of x is going to be relatively small compared to the concentration of HCN. Therefore, we can approximate the equilibrium expression as:
Ka ≈ x2 / 0.75
Solving for x:
x = √(Ka × 0.75) = √(6.17 × 10-10 × 0.75)
This simplifies to x ≈ √(4.6275 × 10-10) = 6.8 × 10-6 M
The pH is the negative logarithm of the H+ concentration:
pH = -log(6.8 × 10-6)
Calculating this value gives a pH of approximately 5.17, which is not one of the answer choices provided. There may have been a miscalculation or incorrect values provided in the options. However, based on the given Ka and the initial concentration of HCN, the pH should be around 5.17.