If the base solution is 0.50 M NaOH, and the volume of the initial acid solution as 25.00 mL, what is the concentration of the acid solution?

Question

CH3COOH(aq)+H2O(l)=CH3COO^-+H3O^+

Equivalence point approximated is :8.3 pH and 12.mL for volume of base

This question comes with a graph:

yaxis is the pH: 0, 2, 4, 6, 8, 10, 12,14,16

xaxis is volume of base(mL): 0,2,4,6,8,10,12,16,18

plotted points: (0,3), (2.3, 4.1), (5, 4.3), (6.1, 4.6), (8,5), (8.6, 5.3), (10, 5.7), (10.4, 5.8),(11.5,5.9), (12,8.3), (12.2,12),(12.13, 12.5)

i hope this description helps

Answer 1

If it took 12 ml of base, then you can find the moles of base, and hence the moles of acid in the 25.0 ml.

Moles base:

12 mls NaOH x 1 L / 1000 mls x 0.50 mols/L = 0.006 mols NaOH

Moles acid:

0.006 mols NaOH x 1 mol CH3COOH / mol NaOH = 0.006 mols CH3COOH

Concentration of CH3COOH:

moles / Liters

0.006 mols / 25.00 ml x 1000 ml / L = 0.24 moles / L = 0.24 M

The pH will be greater than 7 since you are titrating a weak acid with a strong base.