It would produce a solution of acetic acid.
Ch3cooch2ch3 + h2o
CH3COOH + OH- ------> H2O + CH3COO- so it is still an acid plus base gives an acetate salt plus water.
This reaction is:CH3COOCH2CH3 + H2O ↔ CH3COOH + CH3CH2OHThe products are acetic acid and ethanol.
carbon oxygen hydrogen = CH3COOH is the thingy for vinegar
HC2H3O2 + NaOH → NaC2H3O2 + H2O Acetic acid + sodium hydroxide → sodium acetate + water
Ch3coo(-) + h2o ---> ch3cooh + oh(-)
Ch3cooch2ch3 + h2o
its balanced
CH3CH2OOCCH3 + H2O ===> CH3CH2OH + CH3COOH
Dissolving in water = splitting in ionsCH3COONH4 --> CH3COO- + NH4+CH3COO-, acetate is a weak base: CH3COO- + H2O CH3COOH + OH-NH4+, ammonium is a weak acid: NH4+ + H2O NH3 + H3O+Totally in water: CH3COO- + NH4+ CH3COOH + NH3 and 2H2O H3O+ + OH-
CH3COOH + OH- ------> H2O + CH3COO- so it is still an acid plus base gives an acetate salt plus water.
This reaction is:CH3COOCH2CH3 + H2O ↔ CH3COOH + CH3CH2OHThe products are acetic acid and ethanol.
carbon oxygen hydrogen = CH3COOH is the thingy for vinegar
HC2H3O2 + NaOH → NaC2H3O2 + H2O Acetic acid + sodium hydroxide → sodium acetate + water
H^+(aq) + C2H3O2^-(aq) + Na^+(aq) + OH^-(aq) = Na^+(aq) + C2H3O2^-(aq) + H2O(L)Reducing (by crossing out repeated [spectator] ions) gives H^+(aq) + OH^-(aq) = H2O(L)
all are amphoteric ic solutions because in ammonia plus acetic acid case ammonia is base and acetic acid is acid , in next water will behave as base and in last case water will react as acid .
If formed by esterification from pentanol and acetic acid it would be C5H11OH + CH3COOH (plus acid catalyst)------> C5H11OOCCH3 + H2O