Go to a short intro to what Ka is
Go to Solving Ka Problems: Part One
Go to Solving Ka Problems: Part Three
Return to a listing of many types of acid base problems and their solutions
Important note: all constants refered to: Kc, Kw, Ka, and Kb are temperature-dependant. All discussions are assumed to be at 25 °C, i.e. standard temperature.
Here's a second typical weak acid problem:
What is the pH of a 0.300 M solution of benzoic acid? Ka = 6.46 x 10¯5
These are the important equations:
HBz + H2O <==> H3O+ + Bz¯
Ka = ( [H3O+] [Bz¯] ) / [HBz] = 6.46 x 10¯5
Bz¯ refers to the benzoate ion. It is completely unimportant what its formula is.
As before, we want the [H3O+]. So we have:
[H3O+] = xand
[Bz¯] = x
This is because of the one-to-one molar ratio between [H3O+] and [Bz¯] that is created as HBz molecules dissociate.
Remember, the [HBz] started at 0.300 M and went down as HBz molecules dissociated. In fact, due to the one-to-one ratio, it went down by 'x' amount and wound up at an ending value of 0.300 - x.
Next is our 'dropping the subtract x' trick:
0.300 - x approximately equals 0.300
We will check the validity of the trick after completing the calculation. If the approximation exceeds 5%, then we have to use the quadratic.
We now have our equation:
6.46 x 10¯5 = {(x) (x)} / 0.300
which is very easy to solve. We move the 0.300 to the other side to get:
x2 = 1.938 x 10¯5
Taking the square root (of both sides!!), we get:
x = 4.40 x 10¯3 M
Remember:
1) The Ka value is unitless, but x is a molarity.
2) Square root both sides. I have had students square root the x2, but not the other side. Weird, but true.
We finish by taking the pH to get a final answer of 2.356.
Checking for 5% we find we had 1.47%, which is acceptable.
Go to a short intro to what Ka is
Go to Solving Ka Problems: Part One
Go to Solving Ka Problems: Part Three
Return to a listing of many types of acid base problems and their solutions