T Temper888 Full Member 10+ Year Member Joined Sep 5, 2011 Messages 21 Reaction score 13 Jan 28, 2012 #1 Members don't see this ad. How is the equation POH=1/2PKb-1/2log[A-] in Section IV.
MedPR Membership Revoked Removed 10+ Year Member Joined Dec 1, 2011 Messages 18,577 Reaction score 57 Jan 28, 2012 #2 Temper888 said: How is the equation POH=1/2PKb-1/2log[A-] in Section IV. Click to expand... How is it what? Upvote 0 Downvote
Temper888 said: How is the equation POH=1/2PKb-1/2log[A-] in Section IV. Click to expand... How is it what?
mcloaf Full Member 10+ Year Member Joined Jan 21, 2012 Messages 5,173 Reaction score 4,759 Jan 28, 2012 #3 Temper888 said: How is the equation POH=1/2PKb-1/2log[A-] in Section IV. Click to expand... The equation is good. I like to use it. Upvote 0 Downvote
Temper888 said: How is the equation POH=1/2PKb-1/2log[A-] in Section IV. Click to expand... The equation is good. I like to use it.
MedPR Membership Revoked Removed 10+ Year Member Joined Dec 1, 2011 Messages 18,577 Reaction score 57 Jan 28, 2012 #4 mcloaf said: The equation is good. I like to use it. Click to expand... me too. Upvote 0 Downvote
T Temper888 Full Member 10+ Year Member Joined Sep 5, 2011 Messages 21 Reaction score 13 Feb 1, 2012 #5 I meant how do you derive that equation? Upvote 0 Downvote
M MT Headed snow, PBR, and bears Lifetime Donor Verified Member 10+ Year Member Platinum Member Joined May 27, 2011 Messages 1,709 Reaction score 35 Feb 1, 2012 #6 . nvm Upvote 0 Downvote