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: The higher the K a, the stronger the acid

Ka acid ionization constant ◦ – also known as the acid ionization constant. Acid Dissociation Constant

The smaller the value of Ka, the less the acid ionizes in aqueous solution. pH and Ka of a Weak Acid

Ka constants

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

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

% Dissociation % dissociation = [HA] dissociated x 100% [HA] initial

Balanced equation 1. Balanced equation for acid equilibrium Ka 2. Equilibrium constant (Ka) expression ICE Table 3. ICE Table 2. pH of a Weak Acid, Given the Ka

Substitute 4. Substitute the [equilibrium] into the Ka equation. 5. Solve for xi) 100 rule ii) quadratic equation Calculate pH 6. Calculate pH from [H+] pH of a Weak Acid, Given the Ka

Chloracetic acid, is a weak acid (Ka = 1.36 x ). Determine the pH of a 12.0 M solution of chloracetic acid. Try This:

Step 1: [H+] = 10 -pH Step 2: [A-] = [H+] Step 3: 3. Ka, Given the pH and [HA]

You measure the pH of a 0.10 M hypochlorous acid solution, HOCl(aq) and find it to be What is the Ka for hypochlorous acid? Try This:

4. Polyprotic acids A polyprotic acid is capable of donating more than one proton (H 2 CO 3(aq), H 2 SO 4(aq) ) There is an ionization constant for each proton donation (K a1, K a2, etc.) as the ionization occurs in steps. (Table Pg. 803) The K a values become smaller with each ionization step, as the removal of a proton from a negatively charged object becomes more difficult.

4. Polyprotic acids

Polyprotic acids However, the most ionization occurs in the first step. ◦ K a1 >> K a2 > K a3.... Consequently, the [H + ] is predominantly established in the first ionization with the K a1 value. Subsequent ionizations (K a2 & K a3 ) only add minimal amounts of [H + ]. Use K a1 to determine the pH of the solution at equilibrium. Polyprotic acids