Question

If 3.58 mL of a 0.205 M NaOH solution was required to completely neutralize 10.0 mL of acetic acid (CH3COOH). What was the % (m/v) of the acetic acid?

Answer 1

number of mole of NaOH = (molarity of NaOH)*(volume of NaOH in L)
= 0.205*0.00358
= (7.34*10^-4) mole

1 mol of NaOH required 1 mol of CH3COOH
so,
number of mole of CH3COOH = (7.34*10^-4) mole
(mass of CH3COOH)/(molarmass of CH3COOH) = (7.34*10^-4) mole
molarmass of CH3COOH = 60.0 g/mool
(mass of CH3COOH)/60 = (7.34*10^-4) mole
mass of CH3COOH = 0.0441 g

(m/V) % of CH3COOH = {(mass of CH3COOH)/(volume of CH3COOH solution)}*100
= (0.0441/10.0)*100
= 0.441 %

Answer : 0.441 %