(Original post by Asphyxiating)
Thanks, that's really a great help! I found out where I went wrong, with the units
but somehow I get
[H+] = 1.84 * 10^(-5) * 0.011 / 0.039 = 5.1897..x 10(-6)
..instead of 4.717... * 10^(-6)?
I've got Ka as 1.84 x 10(-5) but hey maybe our teacher got it wrong
Anyway thanks! You make it look so easy.. now I know I am definetely going to fail this exam!
It is actually quite simple if you know some equilibrium theory. Il take you through it step by step. First of all, a weak acid dissolved in water discociate to a very small extent. You get an equilibrium in the solution. If HA is an acid, the the general reaction looks like this:
HA <---> (H+) + (A-)
Now, if we considder the equilibrium of this reaction we have an equilibrium constant which, accourding to the definition of th equilibrium constant is:
Ka = [H+][A-] / [HA]
This is simply the formula for the equilibrium constant for the particular reaction.
Suppose now we add a soluable salt, BA, to the solution. The salt discociate almost completely accourding to the equation
BA ---> (B+) + (A-)
Now, since the salt discociate completely, and because the weak acid discociate to such a low extent, almost all the [A-] ions will come from the salt. Therefore the aproximation [A-] = [salt] is a quite an accurate aproximation.
Also, since so few molecules of teh weak acid discociate, we may make another aproximation, namely that [HA] = [acid].
We do in other words have three different equations:
a) Ka = [H+][A-] / [HA]
b) [A-] = [salt]
c) [HA] = [acid].
Substituting b and c into a gives the familiar buffer solution formula:
Ka = [H+][salt]/[acid]
By rearrangeing this equation one obtains:
Ka [acid] / [salt] = [H+]
As you can see, the only chemistry you nedd to knwo is the expression for an equilibrium reaction and that a weak acid only discociate to a small extent. The rest is pure maths.