How many milliliters of 0.100 m naoh are needed to neutralize 50.00 ml of a 0.150 m solution of acetic acid (ch3cooh), a monoprotic acid?

ANSWERS

2016-04-10 05:37:55

The neutralization reaction is NaOH + CH3COOH = Na(+) + CH3COO(-) + H2O At the neutralization point all the CH3COOH will have reacted, so you are only interested in the reactant ratio: 1 mol of NaOH reacts with 1 mol of CH3COOH. The formula of molarity is M = n / V => n = M*V => M1 * V1 = M2 * V2 => V2 = M1 * V1 / M2 => V2 = 50.00 ml * 0.150 M / 0.100 M = 75 ml Answer: 75 ml

ADD ANSWER