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__which means_{3}PO_{4}is the stronger acid__the equilibrium__. We can write an equilibrium constant expression for this:*should*favor the products*K*= $\frac{\left[CH_3COOH\right] \left[H_2PO_{4}^{-} \right] }{ \left[H_3PO_4\right] \left[CH_3COO^{-} \right] }$_{eq} - By multiplying through by [H
^{+}], the following expression is obtained:*K*= $\frac{\left[CH_3COOH\right] \left[H_2PO_{4}^{-} \right] \left[H^{+} \right]}{ \left[H_3PO_4\right] \left[CH_3COO^{-} \right] \left[H^{+} \right]}$_{eq}While: *K*(H_{a}_{3}PO_{4}) = $\frac{ \left[H_2PO_{4}^{-} \right] \left[H^{+} \right]}{ \left[H_3PO_4\right] }$And: *K*(CH_{a}_{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*= K_{eq}_{a }(H_{3}PO_{4}) $*$ $\frac{1}{K_a (CH_{3} COOH)}$ - Simplified further:
*K*= $\frac{K_a(H_3PO_4)}{K_a (CH_{3} COOH)}$_{eq} - 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*= $\large \frac{6.9 \, * \, 10^{-3}}{1.4 \, * \, 10^{-5}}$ = 492.86..._{eq} - 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__.

- For example, if solutions containing CH