Please answer these two questions in detail, I want to learn how to do it. I don
ID: 955087 • Letter: P
Question
Please answer these two questions in detail, I want to learn how to do it. I don't even know what "pH" is so if you could explain that in your answe as well that would be awesome!
(1) Of the following, which is the best base to create a buffer with a pH = 8.0?
Aniline, Kb = 4.3 x 10-10
Hydrazine, Kb = 1.3 x 10-6
Pyridine, Kb = 1.7 x 10-9
Ethylamine, Kb = 6.4 x 10-4
(2) The acid HOBr has a Ka = 2.5x10-9. What is the Kb of OBr- at 25oC?
Please explain how to do these as detailed as possible with explainations, I really need to know how this is done. thanks.
Aniline, Kb = 4.3 x 10-10
Hydrazine, Kb = 1.3 x 10-6
Pyridine, Kb = 1.7 x 10-9
Ethylamine, Kb = 6.4 x 10-4
Explanation / Answer
1) Of the following, which is the best base to create a buffer with a pH = 8.0?
what you must base your selection on is described by the Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
You want pH 8.0. You must then pick an acid with pKa as close to 8.0 as possible. A good rule is that pKa should not be more than 1pH unit away from the desired buffer pH.
But in your equation Kb value is given so you know that
pKa + pKb = 14
so the equation becames
pH = (14-pKb) + log ([salt]/[acid])
so putting the equation pKa = - Log10(Kb)
we got that
Hydrazine, Kb = 1.3 x 10^-6
best fits to the equation and gives the pKa as 8.113
so the correct answer is Hydrazine, Kb = 1.3 x 10-6
2)
The acid HOBr has a Ka = 2.5x10-9. What is the Kb of OBr- at 25oC
As there is no change of temperture during the mixing so we can simply put these value in the equation i.e.
pKa + pKb = 14
=> - Log (Ka) + -Log(Kb) = 14
Putting the value of Ka in the above equation we get
Or in simple formula
Ka * Kb = Kw
=> 2.5 * 10-9 * Kb = 1 * 10-14
=> Kb = 1 * 10-14 / 2.5 * 10-9 = 4 * 10-6
Please ask any doubt if you have always happy to help you