Question

25. In a titration 17ml of 0.2M NaOH was used to neutralize 10ml of acetic acid (CH3 COOH) What is the molarity of the acetic acid?/

Answer 1

The molarity of acetic acid is calculated as follows

find the number of moles of NaOH = molarity x volume
= 17x0.2 = 3.4 moles


write the equation for reaction
CH3COOH + NaOH = CH3COONa +H2O

since ratio is be of CH3COOH to NaOH is 1:1 the moles of CH3COOH is therefore = 3.4 moles

therefore the molarity = moles/volume
= 3.4 /10 = 0.34 M