Gravimetric Method for Pipette Calibration Control

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Gravimetric Method for Pipette Calibration Control

• Set pipette to the desired dispensing volume
• Weigh an empty beaker that will accommodate
• 10 dispensed sampler volumes. Do not touch
• the beaker once it has been weighed.
• Dispense set volume 10 times into the beaker,
• noting weight after each addition.
• Use Deionized, Distilled Water !
• Record temperature of the water and note water
  density as listed in the table.
• Calculate the weight of each aliquot. Record all
  data.
• A- Mean weight of aliquots(g)/Density of
  waterMean Volume(ml)
• B- Mean Volume Expected VolumeVolume Error
• C- Volume Error/Expected Volume x 100 Volume
  Error in
• D- Acceptable Volume Error should be lt3

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• A) How many grams of solid NaOH are required to
  prepare 500 ml of a 0.04 M solution
• B) Express the concentration of this solution in
  terms of N, g/liter, w/v, mg and osmolarity.
• Solution A
• Liters x M number of moles NaOH required
• 0.5 x 0.04 0.02 mole NaOH required
• number of moles Wtg/MW 0.02 Wtg/40 Wtg0.8 g
• Solution B
• M N and the solution is 0.04 N
• 0.8g/500ml solution contains 1.6g/L
• (w/v)g per 100ml 1.6g/L 0.16 g/100ml 0.16
• mg mg per 100 ml 0.16 g/100 160 mg/100ml
  160mg
• NaOH yields two particles (Na and OH-)
  Osmolarity 2 x M 0.08 Osmolar

3

• How many milliliters of 5 M H2SO4 are required to
  make 1500 ml of a 0.002 M H2SO4 solution?
• Solution
• liters x M (dilute solution) liters x M
  (concentrated solution)
• 1.5 x 0.002liters x 5
• 1.5 x 0.002 / 5 0.0006 liters of concentrated
  solution required 0.6 ml

4

• Describe the preparation of 2 liters of 0.4 M HCl
  starting with a concentrated HCl solution (28
  w/w, SG 1.15).
• Solution
• Liters x M number of moles 2 x 0.4 0.8 moles
  HCl needed
• Wtg number of moles x MW Wtg0.8 x 36.5 29.2 g
  pure HCl needed
• The stock solution is not pure but 28 HCl by
  weight so
• 29.2 / 0.28 104.3 g stock solution needed
• instead of weighing out 104.3 g of stock
  solution we can calculate the volume required
• Volume (ml) Wt(g)/? (g/ml) 104.3/1.15 90.7 ml
  stock solution needed

Volume Wt / (? x )
5

• To prepare 1 liter of 0.2 M solution of Acetic
  acid (CH3COOH), how many milliliters of the stock
  solution with the purity of 98 and SG of 1.05
  g/ml are needed?
• Solution
• Volume Wt / (? x )
• Wt(g) needed 0.2 x 60 (MW) 12 g
• Vol. 12/(0.98 x 1.05) 11.66 ml

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