# Acid and Base Dissociation Constants. How do we calculate [H + ] for a weak acid? We know that strong acids dissociate 100% and that, therefore, the [H.

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Acid and Base Dissociation Constants

How do we calculate [H + ] for a weak acid? We know that strong acids dissociate 100% and that, therefore, the [H + ] equals that [acid] that we start with What about weak acids? Don’t ionize 100%, so the [H + ] is NOT the same as our starting concentration of our acid!

Recall K w Autoionization of water: H 2 O ↔ H + + OH - Or H 2 O + H 2 O ↔ H 3 O + + OH - K eq = K w = [H3O + ] [OH - ] = 1.0 x 10 -14

KaKa Similarly to K w we can write an equilibrium expression for the dissociation of a weak acid The equilibrium for a weak, monoprotic acid (HA) looks like this: HA (aq) + H 2 O (l) ↔ H 3 O + (aq) + A - (aq) So, we can write an equilibrium expression that looks like this: where K a is the acid ionization constant

KaKa Example: Write the equilibrium expression for the ionization of acetic acid. CH 3 COOH (aq) ↔ CH 3 COO - (aq) + H + (aq)

K a and Acid Strength K a values are typically between 1 – 1 x 10 -16 The higher the value of K a, the more the acid dissociates in water and, hence, the stronger the acid

What about weak bases? Weak bases also form an equilibrium in water: B (aq) + H 2 O (l) ↔ HB + (aq) + OH - (aq) This can be represented by the base dissociation constant: Like K a, a higher K b means that more B has dissociated and, therefore, the stronger the base

Note: Coefficients and Equilibrium Expressions If you have coefficients in your reaction equation, they become subscripts in the equilibrium expression: 2AB → A 2 + B 2

Try it Write the equilibrium expression for the dissociation of NH 3 in water. NH 3(aq) + H 2 O (l) ↔ NH 4(aq) + + OH - (aq) Try the Self Test 10.2

So how does K a help us find the [H + ]? The K a ’s for almost every weak acid you could think of have been measured (at 25 o C) and recorded If we know the value of K a and the starting concentration of our weak acid, we can solve for [H + ]

Try It: What is the concentration of H + in 0.50M HF at 25 o C? From the acid table, K a = 7.1 x 10 -4, so: HF (aq) ↔ H + (aq) + F - (aq) Now what? Now, we use ICE tables!

ICE table HF (aq) ↔ H + (aq) + F - (aq) Initial (M)0.5000 Change (M)-x+x Equilibrium (M)0.50 – xxx

Solve for x In this case x is our [H + ]

Short Cut: If < 500, the change in the initial concentration (x) is negligible and can be ignored. HF (aq) ↔ H + (aq) + F - (aq) Initial (M)0.5000 Change (M)-x+x Equilibrium (M)0.50 – xxx ]

Percent Dissociation (aka. Percent Ionization) The fraction of molecules that dissociate compared to the initial concentration, expressed as a percent: Percent dissociation = Ex: If a 0.10 M solution of benzoic acid was found to dissociate to give a [H+] = 1.1 x 10-3 M, the percent dissociation would be:

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