0 Members and 1 Guest are viewing this topic.
Oh man, I got it now! my malfunction was confusing BH+ with H+.
The Ka would be [BH+][OH-]/H2O.
BH+ would donate the proton
and Ka would equal [B ][OH-]/[H30] ?
You have the equilibrium B + H2O BH+ + OH-The value of Ka is not of interest here. Just for you to understand, try to set up the equation for Ka. Which equilibrium is described by Ka? (remember conjugated acid/base pairs)
Page created in 0.064 seconds with 21 queries.