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LabFlow

Calculations and Analysis

A. Determination of Initial Concentrations

Initial volume of NaOH: 2.95 mL

Final volume of NaOH: 35.85 mL

Volume of 0.25 M NaOH added (mL) = 35.85-2.95 = 32.9 mL

Initial concentration of acetic acid (M)

Volume of 0.25 M NaOH added 32.9 mL

CH3COOH + NaOH ———> CH3COO-Na+ + H2O

Here, assuming volume of acetic acid = 1 mL

To calculate initial concentration of acetic acid (CH3COOH), using formula equation:

M acetic * V acetic = M naoh * V naoh

M acetic * 1 mL = 0.25 M * 32.9 mL

M acetic = 8.225 M

B. Determination of the Blank

Initial volume of NaOH:2.25 mL

Final volume of NaOH:6.85 mL

(1pts)

Volume of 0.25 M NaOH added (mL) = 6.85 mL – 2.25 mL = 4.60 mL

C. Determination of Final Concentrations

Initial volume of NaOH2.50 mL

Final volume of NaOH16.25 mL

Volume of 0.25 M NaOH added (mL) = 16.25 – 2.50 = 13.75 mL

Volume of 0.25 M NaOH required to neutralize acetic acid (corrected volume, mL) = 13.75 – 4.60 =

9.15 mL

Final concentration of acetic acid (M) = 9.15 x 0.25 / 1 = 2.2875 M

D. Calculation of the Equilibrium Constant

Enter the equilibrium concentrations for all species and calculate the Keq for the reaction.

Equilibrium concentration o …

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