Find moles of base at the equivalence point. Find moles of acid at the equivalen

Question

Find moles of base at the equivalence point. Find moles of acid at the equivalence point. Find mass of acid Find volume of concentrated vinegar Find mass of vinegar solution Find mass percent of acetic acid A vinegar titration was completed using a pH meter as an indicator of the changing pH. A 25.00 mL sample of diluted vinegar (diluted by a factor of 5) was placed in a beaker and subsequently titrated with 0.1098 M NaOH. A derivative curve of the titration suggested that the equivalence point occurred at 40.90 mL. Calculate the mass percent of CH3COOH. IThe density of vinegar is 1.008 gimL)

Explanation / Answer

Ans. Balanced neutralization reaction: CH3COOH + NaOH ------> CH3COONa + H2O

Stoichiometry: 1 mol acetic acid (CH3COOH), the active ingredient of vinegar, is neutralized by 1 mol NaOH.

#1. Moles of Base at equivalence point    

            Moles of NaOH = Molarity x Volume (in L) of NaOH consumed

                                                = 0.1098 M x 0.04090 L                               ; [1 L = 1000 mL]

                                                = (0.1098 mol/ L) x 0.04090 L

                                                = 0.00449082 mol

#2. Moles of Acid at equivalence point: According to the stoichiometry of balanced reaction, 1 mol NaOH neutralized 1 mol acetic acid. Thus, to reach equivalence point, the number of moles of NaOH must be equal to that of moles of acid.

So,

            Moles of acetic acid = moles of NaOH at equivalence point

Hence, Moles of acetic acid = 0.00449082 mol

#3. Mass of acid                                  

            Mass of acid = Moles x molar mass of acid

                                    = 0.00449082 mol x (60.05 g mol-1)

                                    = 0.269673741 g

                                    = 0.27 g

#4. Volume of concentrated vinegar =

Total volume of diluted vinegar x dilution factor

= 25.0 mL x (1/5)

= 5.0 mL

Note: “Dilution by a factor 5” means that 1 mL of concentrated vinegar is diluted to 5.0 mL. so, dilution factor = (1/5) , where 1= vol. of concentrated vinegar ; 5 = total volume of diluted solution.

#5. Since the total acetic acid content of the diluted solution is solely due to the one present in concentrated vinegar,

            Mass of acetic acid in 5.0 mL concentrated vinegar = Mass of acetic acid in 25.0 mL diluted solution

So, Mass of acetic acid in 5.0 mL concentrated vinegar = 0.27 g

#6. Total mass of 5.0 mL concentrated vinegar = Volume x density

                                                            = 5.0 mL x (1.008 g/mL)

                                                            = 5.04 g

We have, mass of acetic acid in this solution = 0.27 g

Now,

            Mas % (wt/ wt) = (mass of acetic acid / mass of solution) x 100

                                    = (0.27 g/ 5.04 g) x 100

                                    = 5.36 %

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.