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?

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

75 ml

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