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Get Started Now- Intro Lesson: a4:18
- Intro Lesson: b4:42
- Intro Lesson: c7:11
- Lesson: 1a2:30
- Lesson: 1b7:03
- Lesson: 24:51

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.

- 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

- Introduction
__Which is the stronger acid / base?__a)Competing conjugate acids/bases.b)Using K_{a}/K_{b}expressions to find K_{eq}.c)Using K_{a}/K_{b}expressions to find K_{eq}(continued). - 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}a)Identify their conjugate bases and write the equilibrium equation for the reaction of the two conjugate pairs.b)Use the K_{a}values above to calculate which side of the equation is favoured. Which is the stronger acid? - 2.
**Use the K**_{a}expression to compare the strengths of two weak acids.

Formic acid, HCOOH, and butanoic acid, CH_{3}CH_{2}CH_{2}COOH, are both weak acids. Their K_{a}acidity constants^{1}are below:

K_{a}(HCOOH): 1.8*10^{-4}

K_{a}(CH_{3}CH_{2}CH_{2}COOH): 1.5*10^{-5}

Write a K_{eq}expression using the two weak acids and their conjugate bases to explain which is the stronger acid.^{1}**Source for acidity constant (K**ATKINS, P. W., & DE PAULA, J. (2006). Atkins' Physical chemistry. Oxford, Oxford University Press._{a}) data:

11.

Acid Base Chemistry and pH

11.1

Introduction to acid-base theory

11.2

Conjugate acids and bases

11.3

Strong and weak acids and bases

11.4

Autoionization of water

11.5

Acid / base dissociation constant (K_{a} and K_{b})

11.6

Relative strengths of acids and bases

11.7

pH, pOH and antilogs

11.8

Mixing strong acids and bases

11.9

Hydrolysis

11.10

K_{a} and K_{b} calculations

11.11

Acid-base titration