To determine the number of moles, multiply the volume in liters by the molarity.

For benzoic acid, V = 0.1 liter

For NaOH, V = 0.05 liter

For benzoic acid, n = 0.1 * 0.280 = 0.028

For NaOH, n = 0.05 * 0.14 = 0.007

https://pubchem.ncbi.nlm.nih.gov/compound/benzoic_...

The formula of benzoic acid is C6H5COOH. According to the drawing at the website above, benzoic acid has one hydroxyl group. This means one mole of benzoic acid will react with one mole of sodium hydroxide. To determine the number of moles of benzoic acid that remain, subtract these two numbers.

n = 0.028 – 0.007 = 0.021

Let’s determine the total volume.

V = 0.1+ 0.05 = 0.15 liter

Molarity = 0.021 ÷ 0.15 = 0.14

Now we can use Ka.

.

C6H5COOH + H2O → C6H5COO-1 + H3O+1

[C6H5COO-1] * [H3O+1] ÷ [C6H5COOH] = 6.4 * 10^-5

The number of moles of the benzoate ion is 0.007.

Molarity = 0.007 ÷ 0.15

This is approximately 0.0467 moles per liter.

(0.007 ÷ 0.15) * [H3O+1] ÷ 0.14 = 6.4 * 10^-5

[H3O+1] = 8.96 * 10^-6 ÷ (0.007 ÷ 0.15) = 1.92 * 10^-4

Now we can determine the pH.

pH = -1 * log 1.92 * 10^-4

The pH is approximately 3.72.

I hope this is helpful for you.

Moles Benzoic acid initially present = 0.1000 L X 0.280 mol/L = 0.0280 mol

Moles NaOH added = 0.050 L X 0.140 mol/L = 7.002X10^-3 mol

Moles benzoic acid remaining = 0.0210 mol

Molarity = 0.0210 mol / 0.150 L = 0.140 M

moles benzoate ion formed = 0.0070 mol

Molarity = 0.0467 M

Ka = 6.4X10^-5 = [H+][Benzoate]/[benzoic aicd]

6.4X10^-5 = [H+](0.0467)/(0.140)

[H+] = 1.92X10^-4 M

pH = 3.72