CH_3COOH + NaOH rightarrow CH_3COONa + H_2O In an acid-base titration of a vineg

Question

CH_3COOH + NaOH rightarrow CH_3COONa + H_2O In an acid-base titration of a vinegar solution with a sodium hydroxide solution, a Bay View chemistry student found that 34.78 mL of a 0.1950 M NaOH solution are need to neutralize 28.65 mL of a vinegar (acetic acid) solution to a phenolphthalein endpoint. Calculate the Molarity of the vinegar (acetic acid) solution. M_A Times V_A = M_B Times V_B (for 1/1 relationships only) Given the values of an acetic acid solution as 0.2335 M and 0.2341 M (from previous titrations) and using the Molarity calculated in problem #1, calculate the average Molarity for the three titrations.

Explanation / Answer

Ans. Part 1: MA VA = MB VB                 - equation 1

            Where, subscript “A = of vinegar”, and “B = of NaOH”

Putting the values in equation 1-

            MA x 28.65 mL = 0.1950 M x 34.78 mL

            Or, MA = (0.1950 M x 34.78 mL) / 28.65 mL

            Hence, MA = 0.2367 M

Therefore, molarity of acetic acid = 0.2367 M

Part 2: Average molarity = (0.2335M + 0.2341M + 0.2367M) / 3

                                    = 0.2347 M

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