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 NaOH to it, then what is the pH of this new solution?

Answer 1

if we take half of solution

mass of NaC2H3O2∙3H2O --> 3.46/2 = 1.73 g

V of HC2H3O2 = 1/2*55 = 27.5 mL

V = 9 mL --> 9/2 = 4.5 mL

Then

mmol of acid = MV = 3*27.5 = 82.5 mmol

mmol of salt (NaC2H3O2∙3H2O = mass/MW = (1.73 )/(136) = 0.01272 mol = 0.0127*10^3 mmol = 12.7 mmol

now...

acid + base = conjguate base

so

mmol of NaOH added = MV = 2*1 = 2 mmol

mmol of acid left = 82.5 -2 = 80.5

mmol of conjgujate formed = 12.7 + 2 = 14.7

substitute in pH equation

pH = pKa + log(A-/HA)

pH = 4.75 + log(14.7/80.5)

pH = 4.011