Question

0.2 mol of gaseous HCl is added to 1.0 L of a buffer solution that contains 3.0 M CH₃COOH and 3.0 M CH₃COONa. What is the pH of the solution? Assume that K_a for CH₃COOH is 10^-5. Assume also that the volume of the solution does not change when the HCl is added.

Type your answer in decimal number carrying only one digit after decimal. So you must round up or down at the second digit after decimal.

For example, your answer must be like 1.7, 2.3, 3.4, 4.2, 5.8, 6.6, 7.4, etc.

Answer 1

Solution :-

HCl is the strong acid therefore when the HCl is added to the buffer then it reacts with the conjugate base and forms the acetic acid

Therefore moles of acetic acid will increase and moles of conjugate base will decrease

Lets first calculate the initial moles of acetic acid and acetate ion in the buffer

Moles = molarity x volume

Initial moles of CH3COOH = 3.0 mol per L * 1 L = 3.0 mol

Initial moles of CH3COO- = 3.0 mol per L * 1.0 L = 3.0 mol

Moles of HCl = 0.20 mol

CH3COONa + HCl ---- > CH3COOH + NaCl

Moles of acid after the addition of HCl are calculated as

New moles of CH3COOH = 3.0 mol + 0.20 mol= 3.2 mol

New moles of CH3COO- = 3.0 mol – 0.2 mol = 2.8 mol

Volume is 1 L therefore the molarities will be same as moles of acid and conjugate base

New molarity of CH3COOH = 3.2 mol / 1 L = 3.2 M

New molarity of CH3COO- = 2.8 mol / 1 L = 2.8 M

Now using the Henderson equation we can calculate the pH of the solution

Ka= 10^-5

Pka = -log ka

      = -log [10^-5]

     = 5

pH= pka + log [conj base ]/[acid]

pH= pka + log [CH3COO-]/[CH3COOH]

pH= 5 + log([2.8]/[3.2])

pH= 4.9

Therefore the pH of the buffer will be 4.9