I\'m trying to prepare for a chemistry test. Can someoneplease give me some dire

Question

I'm trying to prepare for a chemistry test. Can someoneplease give me some direction with this problem? A tablet of a common over the counter drug contains 200 mg ofcaffeine(C8H10N4O2). Whatis the pH of the solution resulting from the dissolution of two ofthese tablets in 225 ml of water at 25 deg C? (For caffeine,Kb = 4.1 x 10-4)
A. 2.76 B. 7.67 C. 10.96 D. 6.33 E.11.24

Thank you very much!

Adama

I'm trying to prepare for a chemistry test. Can someoneplease give me some direction with this problem? A tablet of a common over the counter drug contains 200 mg ofcaffeine(C8H10N4O2). Whatis the pH of the solution resulting from the dissolution of two ofthese tablets in 225 ml of water at 25 deg C? (For caffeine,Kb = 4.1 x 10-4)
A. 2.76 B. 7.67 C. 10.96 D. 6.33 E.11.24

Thank you very much!

Adama

Explanation / Answer

We nedd to calculate the solution that contains the 2tablets.We know that each tablet contains 200 mg of Caffiene. Thus the total weight of the Caffiene is present in thesolution = 400 mg Therefore the number of moles of the Caffiene is present= weight/M.Wt                                                                                       =400*10-3 g /194.19g/mole                                                                                      = 2.0*10-3 mole Concentration of the solution = number ofmoles/volume of the solution                                               =0.0020 mole/0.225L                                              = 9.15*10-3 mole/L Therefore [OH-] at equillibrium =Kb*C                                              = 4.1 x 10-4*9.15*10-3                                              = 3.7*10-6 M Since pOH = -log[OH-]                = -log(3.7*10-6)                = 5.42 pH = 14 - pOH        = 14 - 5.42        = 8.58                                              = 3.7*10-6 M Since pOH = -log[OH-]                = -log(3.7*10-6)                = 5.42 pH = 14 - pOH        = 14 - 5.42        = 8.58

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