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

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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 theequivalence point?

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4.3 Ratings (836 Votes)

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^-3Â Â Â Â Â Â Â Â Â Â 4.5 x 10^-3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0------------------> initial

0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.5 x 10^-3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 4 x 10^-3Â Â Â Â Â Â Â 4 x 10^-3---------------> equivalence point

in the solution NaOH remained.

moles of remained solution = 0.5 x 10^-3Â Â

molarity of solution = moles / total volume

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.5 x 10^-3Â Â / 40 +12.76

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 9.47 x10^-6

molarity of solution = 9.47 x10^-6 M

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