How to determine the stronger of two acids/bases in reactions between two competing conjugate pairs.
How to use ka/kb expressions to construct expressions for the equilibrium constant of two competing conjugate pairs.
How to calculate the equilibrium constant with two competing conjugate pairs.
In Acid dissociation constant
we looked at the expression for ka (and kb), which tells us how dissociated a weak acid (or weak base) is in solution. The larger the Ka (or Kb) value, the stronger the acid (or base).
That the Ka value is basically an “acid strength rating” is good to remember for solutions with multiple conjugate pairs. When two conjugate pairs are mixed together in equilibrium, there are two competing acids (and bases) both trying to be an acid – trying to donate H+ (and two competing bases both trying to accept H+). You can use the values for Ka and Kb to see which one is the stronger acid/base, and which side is the equilibrium favors.
For example, if solutions containing CH3COOH and H3PO4 were combined, there would be two acids both competing to donate protons to other species.
The stronger acid will do this to a greater extent. The Ka values1 will identify which it is: Ka (CH3COOH) = 1.4 * 10-5, while Ka (H3PO4) = 6.9 ∗ 10-3. This shows that H3PO4 donates protons with greater ability than CH3COOH.
Therefore an equilibrium can be written:
H3PO4 + CH3COO-⇌ CH3COOH + H2PO4-
The Ka values show, H3PO4 is the stronger acid which means the equilibrium should favor the products. We can write an equilibrium constant expression for this:
Keq = [H3PO4][CH3COO−][CH3COOH][H2PO4−]
By multiplying through by [H+], the following expression is obtained:
Notice these are now expressions of dissociation? This can now be simplified using Ka expressions!
Keq = Ka (H3PO4) ∗Ka(CH3COOH)1
Keq = Ka(CH3COOH)Ka(H3PO4)
Both of these are known constants (we used them earlier!) so we can calculate an equilibrium constant to determine which side of the equilibrium is favored; it should back up our prediction that H3PO4 dissociates more:
Keq = 1.4∗10−56.9∗10−3 = 492.86...
This value (a ratio of around 493:1) shows the equilibrium heavily favors the products as predicted by the acid dissociation constants. For any two conjugate pairs in competition, look up the Ka value for both conjugate acids then set up the equilibrium and Keq expression like this:
HX = conjugate acid (stronger acid; larger Ka),
HY = conjugate acid (weaker acid; smaller Ka),
X- = conjugate base of HX
Y- = conjugate base of HY
If done correctly, the Keq expression will yield a value greater than 1 (showing the equilibrium shifted right; that the stronger acid dissociates more). Another way to read this is that the equilibrium will favor the side with the weaker acid.
Which is the stronger acid / base?
Find the equilibrium constant for two competing weak acids and their conjugate pairs.
Ethanoic acid, CH3COOH and carbonic acid, H2CO3 are both weak acids. Their Ka acidity constants1 are below:
Ka (CH3COOH) = 1.4*10-5 Ka (H2CO3) = 4.5*10-7
Relative strengths of acids and bases
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