The initial sample of vinegar = 25mL (diluted by a factor of 5) was titrated wit

Question

The initial sample of vinegar = 25mL (diluted by a factor of 5) was titrated with .1098 M NaOH Derivative curve shows equivalence point at 40.90 mL Calculate mass percent of CH3COOH (density of vinegar is 1.008g/mL)
I am not sure about the proper way to set this up, any helpwould be appreciated, thanks. was titrated with .1098 M NaOH Derivative curve shows equivalence point at 40.90 mL Calculate mass percent of CH3COOH (density of vinegar is 1.008g/mL)
I am not sure about the proper way to set this up, any helpwould be appreciated, thanks.

Explanation / Answer

Equivalence volume of NaOH = 40.9 mL Molarity = 0.1098 M Moles of NaOH = molarity * volume (L)                          = 0.1098M * 0.0409 L                          = 0.0045moles . Since acetic acid and sodium hydroxide react in a 1:1 ratio, moles of acetic acid =0.0045 moles mass of acetic acid = moles * molar mass                             = 0.0045 moles * 60.05 g/mol                             = 0.27 g . Volume of solution = 25 mL Density = 1.008 g/mL Mass of solution = density * volume                          = 1.008g/mL * 25 mL                          = 25.2g . Mass percent of acetic acid = ( mass of acetic acid / mass ofsolution) * 100                                       = ( 0.27g / 25.2 g) * 100                                       = 1.1%

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