acid base equilibrium

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Acid Base Equilibrium Weak Acids & Bases

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Acid Base Equilibrium. Weak Acids & Bases. Recall From Yesterday…. pH = -log [H 3 O + ] [H 3 O + ] = 10 -pH pOH = -log [OH - ] [OH - ] = 10 -pH pK w = pH + pOH = 14. Weak Acids. Donate only one H + Do not dissociate at 100% For a Weak Acid: - PowerPoint PPT Presentation

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Page 1: Acid Base Equilibrium

Acid Base Equilibrium

Weak Acids & Bases

Page 2: Acid Base Equilibrium

Recall From Yesterday….

pH = -log [H3O+]

[H3O+] = 10-pH

pOH = -log [OH-][OH-] = 10-pH

pKw = pH + pOH = 14

Page 3: Acid Base Equilibrium

Weak AcidsDonate only one H+

Do not dissociate at 100%

For a Weak Acid:

The higher the Ka, the stronger the acid

Page 4: Acid Base Equilibrium

Ka ◦– also known as the acid ionization constant.

Acid Dissociation Constant

Page 5: Acid Base Equilibrium

The smaller the value of Ka, the less the acid ionizes in aqueous solution.

pH and Ka of a Weak Acid

Page 6: Acid Base Equilibrium

Ka constants

Page 7: Acid Base Equilibrium

the fraction of acid molecules that dissociate compared with the initial concentration of the acid, expressed as a percent.

depends on the value of Ka for the acid, as well as the initial concentration of the weak acid.

For a Weak Acid:

Or

1. % Ionization

Page 8: Acid Base Equilibrium

A 0.25M solution of HF(aq) is ionized at 21.3%, calculate the pH

Try This:

Page 9: Acid Base Equilibrium

% Dissociation

% dissociation = [HA]dissociated x 100%

[HA] initial

Page 10: Acid Base Equilibrium

1. Balanced equation for acid equilibrium

2. Equilibrium constant (Ka) expression

3. ICE Table

2. pH of a Weak Acid, Given the Ka

Page 11: Acid Base Equilibrium

4. Substitute the [equilibrium] into the Ka equation.

5. Solve for x i) 100 ruleii) quadratic

equation

6. Calculate pH from [H+]

pH of a Weak Acid, Given the Ka

Page 12: Acid Base Equilibrium

Chloracetic acid, is a weak acid (Ka = 1.36 x 10-3).

Determine the pH of a 12.0 M solution of chloracetic acid.

Try This:

Page 13: Acid Base Equilibrium

Step 1: [H+] = 10-pH

Step 2: [A-] = [H+]

Step 3:

3. Ka, Given the pH and [HA]

Page 14: Acid Base Equilibrium

You measure the pH of a 0.10 M hypochlorous acid solution, HOCl(aq) and find it to be 4.23.

What is the Ka for hypochlorous acid?

Try This:

Page 15: Acid Base Equilibrium

4. Polyprotic acids

A polyprotic acid is capable of donating more than one proton (H2CO3(aq), H2SO4(aq))

There is an ionization constant for each proton donation (Ka1, Ka2, etc.) as the ionization occurs in steps. (Table Pg. 803)

The Ka values become smaller with each ionization step, as the removal of a proton from a negatively charged object becomes more difficult.

Page 16: Acid Base Equilibrium

4. Polyprotic acids

Page 17: Acid Base Equilibrium

Polyprotic acids

However, the most ionization occurs in the first step.◦Ka1>> Ka2 > Ka3 .. . .

Consequently, the [H+] is predominantly established in the first ionization with the Ka1 value. Subsequent ionizations (Ka2 & Ka3) only add minimal amounts of [H+].

Use Ka1 to determine the pH of the solution at equilibrium.

Polyprotic acids

Page 18: Acid Base Equilibrium