Labflow

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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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