A 25.0 mL sample of vinegar was analyzed via titration. The standardized solutio

Question

A 25.0 mL sample of vinegar was analyzed via titration. The standardized solution of NaOH was 0.70 M. A total of 15.3 mL of the sodium hydroxide were required to reach the pink endpoint. What was the concentration of the acetic acid in the vinegar solution?

Update : the question above it #2.

3. Determine how many moles of acetic acid would be present in 500.0 mL of the vinegar soltuion you found the concentration of in question 2.

4.Determine how many grams of acetic acid are present in the number of moles you found in question 3.

5. A vinegar solution has a density of 1.005 g/mL. How many grams are present in 500.0 mL of a vinegar solution?

Update 2: 6. Using your answers from questions 4 and 5, determine the mass percent of acetic acid in vinegar.

Update 3: for #2 I got 0.0257 g/ml does this seem right?

Explanation / Answer

2. moles NaOH = 0.0153 L X 0.70 mol/L = 0.0107 mol NaOH
Since acetic acid and NaOH react in a 1:1 ratio, the vinegar sample contained 0.0107 moles acetic acid
Concentration = 0.0107 mol / 0.025 L = 0.428 mol/L

(The question doesn't say to express it in g/L, but: 0.428 mol/L X 60.0 g/mol = 25.7 g/L or 0.0257 g/mL)

3. 0.428 mol/L X 0.500 L = 0.214 mol

4. 0.214 mol X 60.0 g/mol = 12.84 grams

5. 500 mL X 1.005 g/mL = 502.5 grams (your answer is fine)

6. (12.84 g / 502.5 g) X 100 = 2.48% (w/w)

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