Posted by HELPP on Wednesday, October 26, 2011 at 5:20pm.
The problem asks for (H^+) and I see 2.9E-12 on my calculator. You must be punching in the wrong numbers.
so i do the -log(3.4x10^-3)=2.46
No, you are making two errors.
#1. The problem asks for (H^+) and gives you the (OH^-). So
(H^+)(OH^-) = Kw = 1E-14
(H^+) = 1E-14/(OH^-) = 1E-14/(3.4E-3) = 2.94E-12 which rounds to 2.9E-12 to two significant figures. That's the answer.
#2. You are using the value of OH as the value for H^+ AND you're changing that to pH which it doesn't ask for.
ok thanks
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