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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Q: A 10.0 mL sample of fruit juice is titrated with 7.55mL of NaOH solution. The molar concentration of…

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Q: Determine the pH during the titration of 58.3 mL of 0.310 M acetic acid (K, = 1.8×10) by 0.310 M…

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Q: A 22.00 mL sample of HCl is titrated to phenolphthalein end point with 24.63 mL of a 0.08481 M NaOH…

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Precipitation Titration

Precipitation analysis is used to analyze drugs, and to measure the salt content in food and beverages. It is an important concept from the examination point of view many practical questions are asked on the topic. This topic is significant in the professional exams for both undergraduate and graduate courses, especially for