Relative strengths of acids and bases  AcidBase Theory
Relative strengths of acids and bases
Lessons
Notes:
In this lesson, we will learn:
 How to determine the stronger of two acids/bases in reactions between two competing conjugate pairs.
 How to use k_{a}/k_{b} expressions to construct expressions for the equilibrium constant of two competing conjugate pairs.
 How to calculate the equilibrium constant with two competing conjugate pairs.
Notes:
 In Acid dissociation constant
we looked at the expression for k_{a} (and k_{b}), which tells us how dissociated a weak acid (or weak base) is in solution. The larger the K_{a} (or K_{b}) value, the stronger the acid (or base).
That the K_{a} 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 K_{a} and K_{b} to see which one is the stronger acid/base, and which side is the equilibrium favors. For example, if solutions containing CH_{3}COOH and H_{3}PO_{4} 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 K_{a} values^{1} will identify which it is: K_{a} (CH_{3}COOH) = 1.4 * 10^{5}, while K_{a} (H_{3}PO_{4}) = 6.9 $*$ 10^{3}. This shows that H_{3}PO_{4} donates protons with greater ability than CH_{3}COOH.
Therefore an equilibrium can be written:H_{3}PO_{4} + CH_{3}COO^{} $\rightleftharpoons$ CH_{3}COOH + H_{2}PO_{4}^{}
The K_{a} values show, H_{3}PO_{4} is the stronger acid which means the equilibrium should favor the products. We can write an equilibrium constant expression for this:K_{eq} = $\frac{\left[CH_3COOH\right] \left[H_2PO_{4}^{} \right] }{ \left[H_3PO_4\right] \left[CH_3COO^{} \right] }$  By multiplying through by [H^{+}], the following expression is obtained:
K_{eq} = $\frac{\left[CH_3COOH\right] \left[H_2PO_{4}^{} \right] \left[H^{+} \right]}{ \left[H_3PO_4\right] \left[CH_3COO^{} \right] \left[H^{+} \right]}$ While: K_{a} (H_{3}PO_{4}) = $\frac{ \left[H_2PO_{4}^{} \right] \left[H^{+} \right]}{ \left[H_3PO_4\right] }$ And: K_{a} (CH_{3}COOH) = $\frac{ \left[CH_3COO^{} \right] \left[H^{+} \right]}{ \left[CH_{3}COOH\right] }$  Notice these are now expressions of dissociation? This can now be simplified using K_{a} expressions!
K_{eq} = K_{a }(H_{3}PO_{4}) $*$ $\frac{1}{K_a (CH_{3} COOH)}$  Simplified further:
K_{eq} = $\frac{K_a(H_3PO_4)}{K_a (CH_{3} COOH)}$  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 H_{3}PO_{4} dissociates more:
K_{eq} = $\large \frac{6.9 \, * \, 10^{3}}{1.4 \, * \, 10^{5}}$ = 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 K_{a} value for both conjugate acids then set up the equilibrium and K_{eq} expression like this:
Where:
HX = conjugate acid (stronger acid; larger K_{a}),
HY = conjugate acid (weaker acid; smaller K_{a}),
X^{} = conjugate base of HX
Y^{} = conjugate base of HYHX + Y^{} $\rightleftharpoons$ HY + X^{} K_{eq} = $\large\frac{[Proucts]}{[Reactamts]}$ or: K_{eq} = $\large\frac{K_a (HX,\; acid\; in\; reactants)}{K_a (HY,\; acid\; in\; products)}$  If done correctly, the K_{eq} 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.

Intro Lesson
Which is the stronger acid / base?

1.
Find the equilibrium constant for two competing weak acids and their conjugate pairs.
Ethanoic acid, CH_{3}COOH and carbonic acid, H_{2}CO_{3} are both weak acids. Their K_{a} acidity constants^{1} are below:
K_{a} (CH_{3}COOH) = 1.4*10^{5}
K_{a} (H_{2}CO_{3}) = 4.5*10^{7}