calculate the concentration of sodium benzoate that must be present in a 2.0M solution of benzoic acid (C6H5COOH) to produce a pH of 4.00.

THIS IS THE ONLY INFORMATION GIVEN. I CANNOT FIND Kb ANY WHERE IN MY TEXT NOR DOES IT GIVE US A VOLUME. CAN ANYONE HELP?

To calculate the concentration of sodium benzoate in a 2.0 M solution of benzoic acid to achieve a pH of 4.00, we need to make some assumptions and use the information available.

First, let's assume that the benzoic acid (C6H5COOH) will fully dissociate into hydrogen ions (H+) and benzoate ions (C6H5COO-) in water. In this case, we can consider the concentration of benzoic acid as the concentration of hydrogen ions (H+).

By using the pH value of 4.00, we can calculate the concentration of H+ ions using the pH equation:

pH = -log[H+]

Rearranging the equation:

[H+] = 10^(-pH)

[H+] = 10^(-4.00)

[H+] = 1.0 x 10^(-4) M

Now, since we assumed that the benzoic acid fully dissociates, the concentration of benzoate ions (C6H5COO-) will also be 1.0 x 10^(-4) M.

Finally, since sodium benzoate dissolves completely into sodium ions (Na+) and benzoate ions (C6H5COO-) in water, the concentration of sodium benzoate will also be 1.0 x 10^(-4) M.

Therefore, the concentration of sodium benzoate that must be present in a 2.0 M solution of benzoic acid to produce a pH of 4.00 is 1.0 x 10^(-4) M.