Answer to Question #74656 in General Chemistry for Rachel

Question

1. Using the pipette transfer 2mls vinegar into three 250 ml conical flasks.

2. To each flask add about 75mls of water and 3 drops of phenophthalein indicator

3. Using the beaker provided, carefully fill the burette provided with 0.1M sodium hydroxide (NaOH)

4. Slowly add (titrate) 0.1M sodium hydroxide (NaOH) solution from the burette into the flask, swirling gently.

5. Continue adding the NaOH until the end point is reached

6. Record the volume of NaOH added to neutralise the vinegar

*Explain the process of titration in this experiment, writing the equation for what is happening during the titration*

Answer 1

Vinegar is a common household item containing acetic acid as well as some other chemicals.

CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)

By adding the sodium hydroxide, which is a basic solution, to the acetic acid, which is an acidic solution, a neutralization reaction occurs. An indicator known as phenolphthalein, is also added to the vinegar. This indicator turns the solution to a dark pink when excess NaOH is added to make the solution more basic. Thus, when the solution turns light pink, this indicates that it has successfully been neutralized. The amount of NaOH used to standardize the vinegar can then be used to determine the amount of acetic acid in the vinegar as they are both at a 1:1 ratio as seen in the above equation. Thus, the moles of NaOH used to neutralize the acid must equal the number of the moles of acetic acid present in the vinegar.

Regular household vinegar is about 5% acetic acid dissolved in water.

С(CH3COOH)*V(CH3COOH) = C(NaOH)*V(NaOH)

Density of vinegar = 1.006 g/mol

Mass of vinegar = 2 ml*1.006g/ml = 2.012 g

m(CH3COOH) = 2.012*0.05 = 0.1006 g

n(CH3COOH) = m/M = 0.1006g / 60g/mol = 0.00168 mol

C(CH3COOH) = 0.00168mol / 0.077L = 0.0218M

0.0218mol/L*0.077L = 0.1mol/L*V(NaOH)

V(NaOH) = 0.0218*0.077/0.1 = 0.0168L = 16.8 mL