Pregunta: A student titrated 25.00 mL of a vinegar solution with density = 1.00 g/mL. The acetic acid in the vinegar required 39.50 mL of 0.500 M NaOH to reach the end point of the reaction. phenolphthalein titration. The acid-base neutralization reaction is: CH3COOH(aq) + NaOH(aq)  CH3COONa (aq) + H2O(l) Calculate the % w/w of acetic acid in the vinegar.

A student titrated 25.00 mL of a vinegar solution with density = 1.00 g/mL. The acetic acid in the vinegar required 39.50 mL of 0.500 M NaOH to reach the end point of the reaction.
phenolphthalein titration. The acid-base neutralization reaction is:
CH3COOH(aq) + NaOH(aq)  CH3COONa (aq) + H2O(l)

Calculate the % w/w of acetic acid in the vinegar.

traduccion

Un estudiante tituló 25.00 mL de una solución de vinagre con una densidad = 1.00 g/mL. El ácido acético en el vinagre requirió 39.50 mL de NaOH 0.500 M para llegar al punto final de la titulación con fenolftaleína. La reacción de neutralización ácido-base es: CH3COOH(aq) + NaOH(aq)  CH3COONa (aq) + H2O(l) Calcule el % peso/peso del ácido acético en el vinagre.