# A vinegar sample contained 4.60% acetic acid. How many mL of 0.450/V sodium hydroxide would be required?

to titrate 25.00 mL of vinegar sample? Assume that the density of the sample to be 1.00 g mL-1

4.60 g of CH3COOH in 100 g or 100 mL ( since d= 1.00)
MM = 60 g/mol
4.60 / 60 = 0.0767 moles CH3COOH
[CH3COOH ] = 0.0767 / 0.100 L = 0.767 M
moles in 25 mL = 25 x 0.767 /1000 = 0.0192
Moles NaOH needed = 0.0192
V = moles / M = 0.0192 / 0.450 = 0.0426 L = 42.6 mL
I supposed that [NaOH] = 0.450 M

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