Astudent carried out a titration using hc2h3o2(aq) and naoh(aq). the net ionic equation for the neutralization reaction that occurs during the titration is represented above. the naoh(aq) was added from a buret to the hc2h3o2(aq) in a flask. the equivalence point was reached when a total of 20.0ml of naoh(aq) had been added to the flask. how does the amount of hc2h3o2(aq) in the flask after the addition of 5.0ml of naoh(aq) compare to the amount of hc2h3o2(aq) in the flask after the addition of 1.0ml of naoh(aq), and what is the reason for this result?

The  ratio of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) to HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq) is 15 : 19 .

Explanation:

HC2H3O2 is  CH₃⁻ COOH, which is also known as Acetic acid.

IUPAC name of this compound is Ethanoic acid.

Acetic acid has a basicity of 1. so there is one acidic hydrogen is acetic acid.

Given that, equivalence point was reached when 20.0mL of NaOH is added.

let the normality of acetic acid is N₁ and that of NaOH is N₂.

    volume of acetic acid is V₁ and that of NaOH is V₂.  

 Equivalence point occurs when, N₁ × V₁ =  N₂ × V₂.

⇒  N₁ × V₁ =  N₂ × 20.

after the addition of 5.0mL of NaOH(aq), remaining N₁ × V°  = N₂ × (20 - 5).

                                                                                                  = N₂ × 15.

after the addition of 1.0mL of NaOH(aq), remaining N₁ × Vˣ = N₂ × (20 - 1).

                                                                                                  = N₂ × 19.

⇒ V° : Vˣ = 15 : 19 .

⇒

i just learnt this recently but i need to look through my notes again