In: Chemistry
How do you calculate the molarity of 40 mL of 0.1M acetic acid (HC2H3O2) solution if 12.76 mL 0.35 NaOH was added to reach the equivalence point?
moles of acetic acid = 0.1 x 40 / 1000 = 4 x 10^-3
moles of NaOH = 0.35 x 12.76 / 1000= 4.5 x 10^-3
CH3COOH + NaOH -----------------------> CH3COONa + H2O
4 x10^-3 4.5 x 10^-3 0 0------------------> initial
0 0.5 x 10^-3 4 x 10^-3 4 x 10^-3---------------> equivalence point
in the solution NaOH remained.
moles of remained solution = 0.5 x 10^-3
molarity of solution = moles / total volume
= 0.5 x 10^-3 / 40 +12.76
= 9.47 x10^-6
molarity of solution = 9.47 x10^-6 M