Asked by JM
If a 0.50 M solution of potassium hypochlorite, [OH^-]=3.0*10^-4. What is the Ka for hypochlorous acid?
Can someone please help me with this question? I can't get the right answer
Can someone please help me with this question? I can't get the right answer
Answers
Answered by
DrBob222
..............OCl- + HOH ==> HOCl + OH-
initial.......0.5.............0......0
change........-x.............+x......+x
equil........0.5-x............x.......x
x = (OH-) = 3.0E-4
Kb = (Kw/Ka) = ((OH-)^2/(OCl-)
Ka = (Kw)(OCl^-)/(OH-)^2
Ka = (1E-14)(0.5)/(3.0E-4)^2 = 5.56E-8
I have a 15 year old text that lists Ka as 3.0E-8.
initial.......0.5.............0......0
change........-x.............+x......+x
equil........0.5-x............x.......x
x = (OH-) = 3.0E-4
Kb = (Kw/Ka) = ((OH-)^2/(OCl-)
Ka = (Kw)(OCl^-)/(OH-)^2
Ka = (1E-14)(0.5)/(3.0E-4)^2 = 5.56E-8
I have a 15 year old text that lists Ka as 3.0E-8.
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