Still Confused?

Try reviewing these fundamentals first

Still Confused?

Try reviewing these fundamentals first

Nope, got it.

That's the last lesson

Start now and get better math marks!

Get Started NowStart now and get better math marks!

Get Started NowStart now and get better math marks!

Get Started NowStart now and get better math marks!

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 Theory

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 and base dissociation constant (Ka and Kb)

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

Ka and Kb calculations