CHM 151L Laboratory Practical

Example Problems

1.         An NaOH solution was prepared by diluting 25.00mL of 2.136 M NaOH to a final total volume of 500 mL. The dilute NaOH solution was then used to titrate three samples of the unknown. The sample mass and volume of diluted NaOH used for each sample were: 0.340 g and 12.34 mL; 0.673 g and 24.58 mL; and 0.417 g and 12.54 mL. What is the exact concentration of the dilute NaOH and the percent by mass of KHSO₄ in the unknown?

NaOH + KHSO₄ à NaKSO₄ + H₂O

Moles of NaOH = Moles of KHSO₄ at end point

V1 = 25.00 mL                                                            Trial 1 = 0.340 g and 12.34 mL

M1 = 2.136 M                                                             Trial 2 = 0.673 g and 24.58 mL

V2 = 500 mL                                                               Trial 3 = 0.417 g and 12.54 mL

Concentration of the dilute NaOH:        V1M1 = V2M2           

Percent by Mass KHSO₄:        

Trial 1:

Trial 2: 53.1%              Trial 3: 43.7%              ß Same Calculations as #1

Don’t use average! Use Median = 52.8%

Please Note that in problem #1 using the average % of KHSO₄ for the 3 trials instead of the median will result in a much lower answer, 49.9% compared to 52.8%.Since trial #3 is much lower than trials 1&2 an error was most likely made in #3. The median gives less weight to such values as #3.

2.         What glassware from your locker could have been used to prepare 100 mL of the dilute NaOH solution for Question One?

In order to prepare 100 mL of dilute NaOH, you should use your 100 mL volumetric flask and a volumetric pipet. This glassware will give you the best accuracy! The size of volumetric pipet can be determined as follows: