Question

An acetic acid/ sodium acetate buffer solution similar to the one you made in the lab was prepared using the following components:

3.46 g of NaC₂H₃O₂∙3H₂O (FW. 136 g/mol)

9.0 mL of 3.0 M HC₂H₃O₂

55.0 mL of water

If you take half of this solution and add 2 mL of 1.00 M HCl to it, then what is the pH of this new solution?

Answer 1

moles of acetate = 3.46g/136 g/mol

= 0.02544 mol

moles of acetic acid = 9mL x 3M /1000

= 0.027 mol

total volume of solution 55+9 = 64 mL

Now the used buffer is half of this solution

that is 32 ml volume with

acetate = 0.02544/2 =0.01272 mol

acid = 0.027/2 = 0.0135mol

now the buffer is

A - + H+ ------------------> HA

0.01272 0 0.0135 initial moles

- 0.002 - change

0.01072 0 0.0155 after reaction

The pH of this buffer is given by Hendersen equation as

pH = pKa + log [A-]/[HA}

= 4.75 + log 0.01072 /0.0155

= 4.599