The acid dissociation Ka of benzoic acid C6H5CO2H is ×6.310−5.Calculate the pH of a ×7.910−4M aqueous solution of benzoic acid. Round your answer to 2 decimal places.

To calculate the pH of a solution of benzoic acid, we can use the formula for the acid dissociation constant (Ka) and the definition of pH.

Step 1: Write the expression for the acid dissociation constant (Ka) Ka = [C6H5CO2-][H3O+] / [C6H5CO2H]

Step 2: Assume that the initial concentration of benzoic acid ([C6H5CO2H]) is x and that it dissociates into x moles of C6H5CO2- and H3O+.

So, Ka = xx / (7.9110^-4 - x)

Given that Ka = 6.3*10^-5, we can set up the equation:

6.310^-5 = x^2 / (7.9110^-4 - x)

Step 3: Since Ka is very small, the amount of acid that ionizes is also very small. Therefore, we can make the assumption that x is much smaller than 7.91*10^-4, so the equation simplifies to:

6.310^-5 = x^2 / 7.9110^-4

Step 4: Solve for x, which represents the concentration of H3O+ ions:

x = sqrt(6.310^-5 * 7.9110^-4) = 7.05*10^-5 M

Step 5: Use the definition of pH to find the pH of the solution:

pH = -log[H3O+]

pH = -log(7.05*10^-5) = 4.15

So, the pH of a 7.91*10^-4 M aqueous solution of benzoic acid is approximately 4.15.