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

moles of sodium acetate= mass/molar mass= 3.46/136=0.0254, moles of acetic acid= molarity* volume in L= 3*9/1000= 0.027 moles, water added= 55ml, acetic acid= 9ml, total volume= 55+9=64 ml

volume considered is 50% of the solution volume= 32 ml.

moles present in 32 ml of solution : sodium acetate =0.0254/2= 0.0127 and acetic acid= 0.027/2=0.0135

moles of NaOH added= 1*2/1000 =0.002, NaOH reacts with NaOH as CH3COOH+ NaOH------>CH3COONa+ H2O

1 mole of acetic acid reacts with 1 mole of NaOH to give 1 mole of sodium acetate. However, moles of NaOH is limited at 0.002 and hence this will be the limiting reactant and moles of sodium acetate formed= 0.002

moles of total sodium acetate= 0.0127+0.002=0.0147, moles of acetic acid remaining = 0.0135-0.002= 0.0115

volume of solution after mixing= 32+2= 34 ml

concentrations :Acetic acid =0.0115/(34/1000) and sodium acetate= 0.0147/(34/1000)

pH= pKa+ log [sodium acetate]/[Acetic acid] =4.76+ log (0.0115/0.0147)= 4.653