Vinegar is a dilute acetic acid solution. Write an equation for acetic acid dissociation $(H_3O^+$ will be simplified by $H^+)$.
$CH_3COOH$ $\rightarrow $ $CH_3COO^-$ $+$ $H^+$.
The reaction between acetic acid and sodium hydroxide is a neutralization reaction. Identify the reagents!
Reagents: Hydrogen ion $H^+$ and hydroxide ion $OH^-$.
Establish the equation involving only these two reagents!
$H^+$ $+$ $ OH^-$ $\rightarrow$ $ H_2O $
Write a nonionic simplified equation!
$CH_3COOH$ $+$ $NaOH$ $\rightarrow$ $ H_2O$ $+$ $CH_3COONa$ In the absence of phenolphthalein nothing is observed, since the solutions of the two reagents are colorless and the product solution also. Phenolphthalein turns purple as soon as there remain $OH^-$ ions which can no more react because all the acid was consumed (equivalence point)
Calculate the percentage of vinegar (d = 1) knowing that $10\; mL$ were neutralized by exactly $40$ mL NaOH $0,25\frac{mol}{L}$ ?
At the equivalence point: $n_{CH_3COOH}$ initialement present = $n_{NaOH}$ added $n_{CH_3COOH}$ $=$ $V_{NaOH}[NaOH]$ = $0.04\cdot 0.25$ $=$ $0.01\; mol$ $m_{CH_3COOH}$ $=$ $n_{CH_3COOH}\cdot M_{NaOH}$ = $0.01\cdot 60$ $=$ $0.6 \;g$ Since $d=1$, the mass of vinegar amounts to $10 \;g$: $ \%_{CH_3COOH}$ = $\frac{0.6\cdot 100}{10}$ = $6\%$