Question

Acetic acid is an important ingredient of vinegar. A sample of 50.0 mL of a commercial vinegar is titrated against a 1.00 M NaOH solution. What is the molarity of acetic acid present in the vinegar if 5.75 mL of the base are needed for the titration? Write out the complete balanced molecular equation with phase labels What is the complete ionic equation? (Remember to check if strong/weak) What is the net ionic equation?

Answer 1

CH₃COOH _(aq) + NaOH _(aq) CH₃COONa _(aq) + H₂O (l)

In above reaction, NaOH is a strong base. It dissociates completely into ions when dissolved in water. Sodium acetate is a ionic compound , it is also dissociated completely into ions when dissolved in water. Hence, instead of writing NaOH and CH₃COONa we can write them as

NaOH (aq) $\rightarrow$ Na ⁺ (aq) + OH ^- (aq)

CH₃COONa _(aq)$\rightarrow$ Na ⁺ (aq) + CH₃COO ^- (aq)  

Hence, above equation becomes,

CH₃COOH _(aq) + Na ⁺ (aq) + OH ^- (aq)   Na ⁺ (aq) + CH₃COO ^- (aq)  + H₂O (l)

above equation is called as complete ionic equation.

In above equation, Na ⁺ (aq) is present on both sides of equation is called spectator ion. By cancelling them , we can write net ionic equation as

CH₃COOH _(aq) + OH ^- (aq)    CH₃COO ^- (aq)  + H₂O (l)

Calculation of Molarity of acetic acid.

We have formula, M _acid x V _acid = M _base x V _base

Therefore,  M _acid = M _base x V _base / V _acid

M _acid = 1.00 M 5.75 ml / 50.0 ml = 0.115 M

ANSWER : Molarity of acetic acid present in the vinegar = 0.115 M