4) A 25.00-mL sample of vinegar is diluted with deionized to 100.0 mL. Then a 10.00-mL aliquot (portion) of the dilute vinegar solutionis titrated with 0.1055 M NaOH solution using phenolphthalein as theindicator. (a) If 19.60 mL of the base solution is required to reach the end-point, calculate the molar concentration of acetic in dilute vinegar solution. (b) What is the molar concentration of acetic acid in the original (undiluted) vinegar? The acid-base reaction occurs as follows: HC2H3O2(aq) + NaOH(aq) --> H2O(l) + NaC2H3O2(aq)

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4) A 25.00-mL sample of vinegar is diluted with deionized to 100.0 mL. Then a 10.00-mL aliquot (portion) of the dilute vinegar solutionis titrated with 0.1055 M NaOH solution using phenolphthalein as theindicator.

(a) If 19.60 mL of the base solution is required to reach the end-point, calculate the molar concentration of acetic in dilute vinegar solution.

(b) What is the molar concentration of acetic acid in the original (undiluted) vinegar? The acid-base reaction occurs as follows:

HC2H3O2(aq) + NaOH(aq) --> H2O(l) + NaC2H3O2(aq)

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