Recent content by spiff711

Hey guys, I have a quick question. I signed up to take Organic over the summer trying to get in all the pre-reqs for a 2+4 Pharm-D program. I would really like to do well in this class and was wondering is there anything I should review or any good study habits or methods I can use to ace...

Edit- Never mind, I understand. The water dissociates into equal parts so the pH and pOH will both be 6.8. I think at least. Thanks for the help!

Thanks for the response. Well Kw= [H3O] x [OH] So I got the value for pOH from the equation pk = 14 = pOH + pH. I then found the concentrations for H3O and OH by using the formulas for pH (pH = -log [H3O]) and pOH ( pOH= -log[OH]). I then plugged those values into the orginal equation and...

Problem: The pH of pure water at 37°C is 6.80. Calculate Kw, pOH, and [OH-] at this temperature. Work: pH: 6.8 pOH = 14 -6.8 = 7.2 [OH]: 10-7.2 = 6.3 x 10-8 M Kw= [H3O] [OH] = 1.5 x 10-7 x 6.3 x 10-8 = 1x10-14 Real answer: pH:6.8 pOH:6.8 [OH]: 1.6x10-7 Kw: 2.5 x 10-14 What...