I don\'t have a clue on how to solve for the answers I\'ve left blank (there is

Question

I don't have a clue on how to solve for the answers I've left blank (there is no trial 3). Any help would be appreciated. Data and Calculations 1. Measurement of pH and fitration of acetic acid solution Concentration of standardized NaOH titrant O1020 -mol/L Mass concentration of acetic acid 2- g/L Trial 3 Trial 2 Trial 1 3.19 Measured pH of the acetic acid solution Mass of acetic acid solution taken for titration oo, o Initial buret reading of NaOH titrant Final buret reading of NaOH titrant Net volume of NaOH Millimoles of NaOH to end point of titration | 1. o 66 Millimoles of acetic acid in sample Molar concentration of acetic acid solution Calculated molar mass of acetic acid g mL mL mL 10.15 mL 10.1.O m 10.15 mL 10.9 mL mmol mmol I/L mmol mmol mol/L mol/L g/mol g/mol Calculation of the ionization constant for acetic acid from the measured pH of the acetic acid samples. (Average the pH and molar concentration values for the samples of acetic acid you titrated; use the average values in the calculation of the ionization constant.) Average molar concentration of acetic acid for the samples you titrated Average pH Calculate the corresponding [H,0% compute the concentration of A-and HA, and calculate the dissociation constant, K mol/L M A M HA 2. Measurement of the pk, of acetic acid by the half-neutralization method

Explanation / Answer

For Trial 1,

Mass Concentrationo of acetic acid = 2.4 g/L (Just check this once I think its wrong!! Its too less!!)

Mass of acetic acid solution taken for titration = 30.08 g

Volume of acetic acid solution taken for titration = 30.08 / 2.4 = 12.533 L

Moles of acetic acid in sample = 1.066 x 10-3

Volume of titrant NaOH = 10.15 mL = 0.01015 L

Molar Concentration of acetic acid solution = (1.066 x 10-3) / ( 12.533 + 0.01015) = 8.5 x 10-5 M

Molar mass of acetic acid = Mass / Moles = 30.08 / (1.066 x 10-3) = 28217.64 g

For Trial 2,

Mass Concentrationo of acetic acid = 2.4 g/L (Just check this once I think its wrong!! Its too less!!)

Mass of acetic acid solution taken for titration = 30.14 g

Volume of acetic acid solution taken for titration = 30.14 / 2.4 = 12.558 L

Moles of acetic acid in sample = 1.071 x 10-3

Volume of titrant NaOH = 10.20 mL = 0.0102 L

Molar Concentration of acetic acid solution = (1.071 x 10-3) / ( 12.558 + 0.0102) = 8.521 x 10-5 M

Molar mass of acetic acid = Mass / Moles = 30.14 / (1.071 x 10-3) = 28141.92 g

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