If 2.50 mL of vinegar requires 34.90 mL of 0.0960 M NaOH to reach the equivalenc

Question

If 2.50 mL of vinegar requires 34.90 mL of 0.0960 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in 500.0 mL of this vinegar? [density of vinegar = 1.01 g mL^-1] Calculate the mass % of acetic acid in the vinegar in the previous exercise (question#1 above). Assume that the density of the vinegar is 1.01 g mL^-1. If phenolphthalein is used as indicator, what color will the solution be at the endpoint? In your reading, it talks about errors that are possible when doing titrations, what are three of those possible sources of error?

Explanation / Answer

Q1.

V = 2.5 mL of vinegar

V = 34.90 mL of M = 0.096 M of NaOH

find mass of acetic acid in V = 500 mL if D = 1.01 g/mL

so...

first, find mol of base

mmol of base = MV = 0.0960*34.90 = 3.3504 mmol of NAOH

ratio is 1:1

so

1 mol of acid = 1 mol f base

3.3504 mmol of base = 3.3504 mmol of acid

mass = mol*MW = (3.3504*10^-3)(60.05 ) = 0.201191 g of acetic acid

this was in 2,50 mL of sample

so

in 500 mL --> 500/2.5 * 0.201191 = 40.2382 g of acetic acid

Q2.

if we need mass, we need to chagne vineger to mass

mass of vinegar in 500 mL = D*V = 1.01*500 = 505 g of vinegar

% mass = 40.2382 /505* 100 = 7.96796 %

Q3.

The color will start as transparent, since it is acidic, evnetually, when pH = 8-9 approx, the color changes to pink/purple

Q4.

1 - incorrect indicator

2 - excess titrant addition

3 - incorrect meassurement of volumes (ignoring miniscus, etc..)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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