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

Number of mmol of CH3COONa = mass in mg/FW = 3.46 x 1000/136 = 25.44

Number of mmol of CH3COOH = M x V (mL) = 3.0 x 55.0 = 165.0

Number of mmol of CH3COONa in half of solution = 25.44/2 = 12.72

Number of mmol of CH3COOH in half of solution = 165.0/2 = 82.5

Number of mmol of HCl added = M x V (mL) = 1 x 2.0 = 2

Number of mmol of CH3COONa after addition of HCl = 12.72 -2 = 10.72 mmol

Number of mmol of CH3COOH after addition of HCl = 82.5 + 2 = 84.5 mmol

pH = pKa + log [CH3COONa]/[CH3COOH]

pH = 4.75 + log 10.72/84.5

pH = 4.75 + log 10.72 – log 84.5

pH = 4.75 + 1.03 -1.93

pH = 3.85