You are asked to prepare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid (C6H5COOH) and some added sodium benzoate salt (C6H5COONa).

a) Find the pH of the benzoic acid solution BEFORE adding the salt.
b) Use the Henderson-Hasselbalck equation to derive how many moles of the salt you should add.
c) How many grams of the salt should you add?

I don't understand part a)

How to start the process of finding the pH

a) To find the pH of the benzoic acid solution before adding the salt, you can use the equation for the dissociation of benzoic acid:

C6H5COOH ⇌ C6H5COO- + H+

Let's call the initial concentration of benzoic acid [C6H5COOH]0, and the concentrations of the benzoate ion [C6H5COO-]0 and the hydronium ion [H+]0. At equilibrium, the concentrations will be [C6H5COOH], [C6H5COO-], and [H+], respectively.

Since benzoic acid is a weak acid, we can assume that its dissociation is relatively small compared to its initial concentration. Therefore, we can neglect [C6H5COO-]0 compared to [C6H5COOH]0, and we can assume that [H+] = [C6H5COO-] = x. Thus, [C6H5COOH] = [C6H5COOH]0 - x.

Now, we can use the equilibrium expression for the dissociation of benzoic acid:

Ka = [C6H5COO-][H+]/[C6H5COOH]

Where Ka is the acid dissociation constant for benzoic acid.

Since we know the value of Ka for benzoic acid (6.3 x 10^-5), we can substitute the concentrations into the equation and solve for x:

Ka = x^2 / ([C6H5COOH]0 - x)

Simplifying and rearranging the equation:

6.3 x 10^-5 = x^2 / (0.0200 M - x)

This is a quadratic equation that can be solved to find the concentration of [H+]. Once you find [H+], you can calculate the pH using the formula:

pH = -log[H+]

b) To find the number of moles of the salt (C6H5COONa) to be added, we will use the Henderson-Hasselbalch equation. The equation is:

pH = pKa + log([C6H5COO-]/[C6H5COOH])

Rearranging the equation, we get:

pH - pKa = log([C6H5COO-]/[C6H5COOH])

Taking the antilog of both sides:

10^(pH - pKa) = [C6H5COO-]/[C6H5COOH]

Multiplying both sides by [C6H5COOH], we get:

[C6H5COOH] * 10^(pH - pKa) = [C6H5COO-]

Now, we can substitute the known values: pH = 4.00, pKa = -log(Ka) = -log(6.3 x 10^-5), and [C6H5COOH] = 0.0200 M into the equation to find the concentration of [C6H5COO-]. Once you have the concentration, you can convert it to moles by multiplying it by the volume of the solution.

c) To find the mass of the salt (C6H5COONa) to be added, we need to use the molar mass of the salt. The molar mass of sodium benzoate (C6H5COONa) is the sum of the molar masses of its constituent elements: carbon (C), hydrogen (H), oxygen (O), and sodium (Na).

C: 6
H: 5
O: 2
Na: 1

Adding these molar masses, we get the molar mass of sodium benzoate. Then, we can multiply the number of moles of sodium benzoate calculated in part (b) by the molar mass to find the mass of the salt to be added.

Give specific details about what youi don't understand.