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Titration Results: Trial 1 Trial 2 Trial 3 Volume of HCl (pipet) used = 10.00 mL 10.00 mL 10.00 mL NaOH Buret 2nd reading (Vf) = 10.87 mL 21.93 mL 32.90 mL 1st reading (Vi) = 0.00 mL 11.00 mL 22.00 mL Volume of NaOH solution used = (Vf - Vi) 10.87 mL 10.93 mL 10.9 mL Molarity of NaOH = M ///////////////////// ///////////////////// ///////////////////// Molarity of HCl (Show sample calculation below) = V1M1 = V2M2 = .1M 10.00 * M1 = 10.87 * .1 .1087 M .1093 M .109 M Average Molarity of HCl = .1087 + .1093 + .109 / 3 = .109 M Relative deviation (ppt) (Show Sample Calculations below) = ? | xi - x? | / nx * 10³ = In the relative error formula xi is the value of the molarity obtained for a given trial; x? is the average molarity; n is the total number of trials. Calculations: (Show a sample calculation of each unique type of determination for full credit.) Problems: (Show all calculations to obtain full credit.)

          Titration Results:

Trial 1 Trial 2 Trial 3
Volume of HCl (pipet) used = 10.00 mL 10.00 mL 10.00 mL
NaOH Buret 2nd reading (Vf) = 10.87 mL 21.93 mL 32.90 mL
1st reading (Vi) = 0.00 mL 11.00 mL 22.00 mL
Volume of NaOH solution used = (Vf - Vi) 10.87 mL 10.93 mL 10.9 mL
Molarity of NaOH = M ///////////////////// ///////////////////// /////////////////////
Molarity of HCl (Show sample calculation below) = V1M1 = V2M2 = .1M 10.00 * M1 = 10.87 * .1 .1087 M .1093 M .109 M
Average Molarity of HCl = .1087 + .1093 + .109 / 3 = .109 M
Relative deviation (ppt) (Show Sample Calculations below) = ? | xi - x? | / nx * 10³ =

In the relative error formula xi is the value of the molarity obtained for a given trial; x? is the average molarity; n is the total number of trials.
Calculations: (Show a sample calculation of each unique type of determination for full credit.)
Problems: (Show all calculations to obtain full credit.)

Titration Results:

Trial 1 Trial 2 Trial 3
Volume of HCl (pipet) used = 10.00 mL 10.00 mL 10.00 mL
NaOH Buret 2nd reading (Vf) = 10.87 mL 21.93 mL 32.90 mL
1st reading (Vi) = 0.00 mL 11.00 mL 22.00 mL
Volume of NaOH solution used = (Vf - Vi) 10.87 mL 10.93 mL 10.9 mL
Molarity of NaOH = M ///////////////////// ///////////////////// /////////////////////
Molarity of HCl (Show sample calculation below) = V1M1 = V2M2 = .1M 10.00 * M1 = 10.87 * .1 .1087 M .1093 M .109 M
Average Molarity of HCl = .1087 + .1093 + .109 / 3 = .109 M
Relative deviation (ppt) (Show Sample Calculations below) = ? | xi - x? | / nx * 10³ =

In the relative error formula xi is the value of the molarity obtained for a given trial; x? is the average molarity; n is the total number of trials.
Calculations: (Show a sample calculation of each unique type of determination for full credit.)
Problems: (Show all calculations to obtain full credit.)

Added by Thomas C.

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