Question

Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C6H5COOH) and 0.20 M sodium benzoate (C6H5COONa), what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (Ka (C6H5COOH) = 6.5 x 10-5)

Please solve in detail

Answer 1

Solution

Given data

Molarity of benzoic acid = 0.250 M

Molarity of sodium benzoate = 0.20 M

Volume of buffer = 0.250 L

Molarity of HCl = 0.100 M

Volume of HCl = 25.0 ml = 0.025 L

pH of solution after adding HCl = ?

lets first calculate the moles of the each species

moles of benzoic acid = molarity * volume in liter

                                       = 0.250 mol per L * 0.250 L

                                      = 0.0625 mol HCl

Moles of sodium benzoate = 0.20 mol per L * 0.250 L = 0.05 mol

Moles of HCl = 0.100 mol er L * 0.025 L = 0.0025 mol HCl

After the addition of the HCl sodium benzoate ions will combine with H+ of the HCl to form the benzoic acid by the following reaction

C6H5COO- + H+ ----- > C6H5COOH

So moles of the sodium benzoate will decrease by 0.0025

And moles of benzoic acid will increase by 0.0025

So new moles of the benzoic acid = 0.0625 + 0.0025 =0.0650 mol

New moles of sodium benzoate = 0.05 – 0.0025 = 0.0475 mol

Now lets calculate the new molarity at total volume

Total volume = 0.250 L +0.025 = 0.275 L

[benzoic acid ] =0.0650 mol / 0.275 L =0.23636 M

[sodium benzoate] = 0.17272 M

Now lets calculate the pH using the Henderson equation

pH= pka + log ([base]/[acid])

pka = - log ka

pka = - log 6.5E-5

pka=      4.187

now lets put the values in the formula

pH = 4.187 + log [0.17272 /0.23636]

pH= 4.05