CHM1 15 Example 3 Collection

Solution 

First, you need to calculate the number of moles of acetic acid in the original sample.

`m_(CH_3COOH) = 0.3421"g " CH_3COOH" " (1"mol " CH_3COOH)/(60"g "CH_3COOH) = 5.7 * 10^(-3) "mol " CH_3COOH`

At equilibrium,

`m_(CH_3COOH) = m_(NaOH)`          `5.7*10^(-3)"mol " CH_3COOH = 5.7*10^(-3)"mol" NaOH`

Finally, use the number of moles and volume of the titrant to calculate Molarity.

`M_(NaOH) = ("mol " NaOH)/("L " NaOH) = (5.7 * 10^(-3))/(.02703"L") = 0.21 "M "NaOH`