CHM1 15 Example 3 Collection

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on

Aug 4, 2022, 4:14:49 PM

Created by

on

Oct 18, 2019, 1:17:50 AM

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.

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

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