Question

2) Titration of acetic acid using indicator and pH meter Trial 1 oo Mass of empty beaker Mass of beaker + acetic acid solution 51.585 80,863 oo Mass of acetic acid solution oo Volume of acetic acid solution mL

Answer 1

In order to complete the table, you should first subtract the empty beaker mass from the combination of the beaker and the acid, so 80.863 g - 51.585 g = 29.278 g.

Then, to obtain the volume of acetic acid, we should consider the density value of 1.05 g/ml, and then use the formula: $\rho = \frac{m}{V}$ , where we can solve for V as follows:

m 29.2789 P 1.05g/ml = 27.884ml

	Trial 1
Mass Empty Beaker	51.585 g
Mass Beaker + CH3COOH	80.863 g
Mass CH3COOH	29.278 g
	27.884 ml

Now to obtain the titration curve, we should plot the volume of titrate added vs the pH of the solution that is obtained, by which we would have:

00000000000 00 Hd 00000000000000000 0.00 5.00 15.00 20.00 10.00 Vol of NaOH (mL)

the horizontal line that is added shows the approximate value at which the equivalence point is reached, which is the most vertical part of the titration curve. This approximate value is 7.8, which would be your pH and pKa of the acid. The equivalence point of the titration.