Question

How many moles of acetic acid will react with 18.25 ml of 0.7500 M NaOH?

Calculate the mass (in grams) of acetic acid in 10.00 ml of a solution which contains 0.0083 moles of acetic acid.

Assuming that the 10.00 ml sample has an overall mass of 10.00g. Based on the above answer, calculate the percent of this total mass which is due to acetic acid.

Answer 1

HAc + NaOH ---> H2O + NaAc

ratio is 1:1 mol

so

find moles of base

mol base = M*V = 0.75*18.25 = 13.69 mmol of base

therefore there are 13.69 mmol of acetic acid, or 0.01369 mol of Acetic Acid

b)

find mass of Acetic acid in 10 ml of a solution that contains 0.0083 moles of acetic acid

mass = mol*MW = 0.0083*60.05 = 0.4984 grams of Acetic acid

c)

V = 10 ml

m = 10 g. ...

calculate % of mass of acetic acid

% mass = acetic acid/total mass = 0.4984 /10 *100 = 4.98%