Question

Discussion: 1. Compare the molarities of the three samples of vinegar that you analyzed. The "precision" of your titration technique is how similar the three values are (the closer the values, the better your precision). 2. Compare the average molarity you obtained to the molarity of the acetic acid in the vinegar as listed on the vinegar bottle. What could account for any differences between these values? 3. How do your pH values compare to the pH determined using the pH paper? There will probably be a difference between these 2 pH values. What could account for this difference? rgn phth late ACy Hy O,-H+NaoH + Nat+ H20 Potasium KHP PhendPiyla le acidolerles PH=7-5pnk Base Reed lalakons Vinegeracchie add males KHP= moles NaoH Amde Valuun CL) &wlet NADA Ganvect toll)

Answer 1

Let vinegar contains 5% acetic acid (according to commercial one).

i.e. The molarity of acetic acid = 5 g/(60 g/mol) = 0.083333 mol or 83.333 mmol (in 100 mL)

Now, the millimoles of acetic acid in 5 mL vinegar = 5 mL * 83.333 mmol/100 mL = 4.167 mmol

Now, the volume of NaOH (let's say 1 M) required to reach the green end point = 4.167 mmol/(1 mmol/mL) = 4.167 mL