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 0.1055MNaOHsolution using phenolphthalein as theindicator. (a) If 19.60mL 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? At the end-point, the volume of NaOH consumed was found to be 32.00 mL. The acid-base reaction occurs as follows: HC2H3O2(aq) + NaOH(aq) --> H2O(l) + NaC2H3O2(aq)
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Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
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 0.1055MNaOHsolution using phenolphthalein as theindicator.
(a) If 19.60mL 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? At the end-point, the volume of NaOH consumed was found to be 32.00 mL. The acid-base reaction occurs as follows:
HC2H3O2(aq) + NaOH(aq) --> H2O(l) + NaC2H3O2(aq)
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