a) water
pH before mixing = 7
pH after mixing= 12.64
pH change = 5.64
b) 0.148 M NH41+
pH before mixing = ?
pH after mixing= ?
pH change = ?
c) 0.148 M NH3
pH before mixing = ?
pH after mixing= ?
pH change = ?
d) a buffer solution that is 0.148 M in each NH41+ and NH3
pH before mixing = ?
pH after mixing= ?
pH change = ?
I found a but am having a lot of trouble figuring out how to do b c and d. I have no clue.Help please!!! pH nightmare!!?
just keep in mind that when doing dilutions that C1*V1 = C2*V2 so if you are adding 10 ml of 2.78 mol/L NaOH to 630 mL of water and you want to find the new concentration so you can determine the pH, you can just plug in.
(2.78M)(10mL)=(C2)(630mL)
C2 = .004M
with that new conc you can find the pH
you must find the pOH since you know the conc of how much acid you are adding, not base
pOH = -log[OH-]
where [OH-] is the concentration you just solved for
kind in mind that pH + pOH = 14
note, this only works for assumed near 100% ionizationHelp please!!! pH nightmare!!?
uh, yea im not gunna do your homework for you.
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