Ka and pKa?? (1 Viewer)

[benthejack]

hey i was just wondering im studying a solubility section in a practise exam and i found a question which i cant find in my textbook.
it goes like this:
The weak acid HA has Ka=1.20*10^-6
Pka=5.92

Calculate the concentration of all species, other than water, that are present in a 0.05mol/L solution of the acid. Discuss the valadity of any assumptions made.


i dont have any idea what this question is talking about (i probably should have listened in class) please help!!!

 

[mercury]

Ka = acid equilibrium constant
pKa = -log[Ka]

If you happen to be doing industrial chemistry, then you would need to learn what they are, but they are not part of acidic environment i believe.

METHOD I (the long way)

Since you are given a weak acid, therefore it does not dissociate 100%. Suppose x amount of the acid dissociated, then you have

HA <=> H+ + A-

initial concentration(mol/L): [HA] = 0.05 [H+]=[A-] = 0
final concentration (mol/L): [HA] = 0.05 - x [H+]=[A-]= x

Sub your values into the expression

Ka = [H+][A-]/[HA] = x^2 / (0.05 - x)

You can solve this quadratic in x, then find out the final concentration of all species
OR you can make an assumption that the amount of acid dissociated, x, is very small because this is a weak acid with such a small ka, so it's negligible compared to 0.05, ie.

0.05 - x approximates to 0.05

then you just have x^2 = Ka, solve for x and sub in to the final concentrations of all species.


METHOD II

pH = pKa + log([A-]/[HA])
Known as Henderson Hasselbach equation, which is applicable in this situation because we are dealing with a weak acid. This is just derived from your Ka expression and it includes the assumption (in this case) that 0.05 - x approximates to 0.05. If you try derive the equation from your Ka expression (just take logs) you'll see why

CONCLUSION

this question is highly unlikely to be asked in your school years, unless you are taking chemistry NQE OR your teacher feels that you can be challenged and shove the question in your face. This type of chemistry questions is actually part of first yr chem syllabus.