2.857E-11 = (x^2)/(0.120-x)
x= 1.9E-6
F- H2O <--> HF OH-
2.857E-11 = (x^2)/(0.120-x)
x= 1.9E-6
2.857E-11 = (x^2)/(0.120-x)
x= 1.9E-6
1.3510^-11 = x^2 / (0.120-x)
2.857E-11 = (x^2)/(0.120-x)
x= 1.9E-6
2.857E-11 = (x^2)/(0.120-x)
x= 1.9E-6
You need to first find Kb, which can be calculated from Ka.
jacob had the right method, but different Kb
Ka for HF is 3.5E-4
Kb = Kw/Ka = (1E-14)/(3.5E-4) = 2.857E-11
2.857E-11 = (x^2)/(0.120-x)
x= 1.9E-6
This post is last updated on hrtanswers.com at Date : 1st of September – 2022