How do you calculate the molarity of 40 mL of 0.1M acetic acid (HC2H3O2) solution if...

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?

Expert Solution

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

queen_honey_blossom answered 10 months ago

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