Ka – The Acid Dissociation Constant

HCl (aq) + NaOH (aq)  HOH (l) + NaCl (aq) Strong Acids and Bases Recall: strong acids and bases completely dissociate (break apart into ions) when dissolved in water Example: HCl (aq) + NaOH (aq)  HOH (l) + NaCl (aq) is really H+ + Cl- + Na+ + OH-  H+ + OH- + Na+ + Cl- Thus, it is easy to calculate the pH or pOH of these solutions.

Weak Acids and Bases Ka: the acid dissociation constant a measure of how much an acid breaks apart into its ions it is similar to the equilibrium constant Keq

CH3COOH + H2O  H3O+ + CH3COO- Example #1 CH3COOH + H2O  H3O+ + CH3COO- Ka = [H3O+][CH3COO-] [CH3COOH] This Ka value now tells us the tendency of CH3COOH to ionize in water.

Ka Values Large Ka value = stronger acid (more ionization) Example: Ka of HF = 6.8 x 10-4 Small Ka value = weaker acid (less ionization) Example: Ka of HCN = 6.2 x 10-10

What do we use Ka values for? Since weak acids only partially dissociate, the calculation of pH using –log [H+] is not accurate. Thus, Ka is used to determine the pH of weak acids. You can also use the pH of weak acids to calculate Ka.

Determining Ka – Example #1 Determine the Ka for a 0.10 M solution of HCOOH if [H3O+] = 4.2 x 10-3 M.

Determining Ka – Whiteboards Find the Ka of a 2.00 M HClO2 solution if [H3O+] = 0.14 M. In a 0.50 M solution of a weak acid HX, the [H3O+] = 8.0 x 10-2 M. Calculate the acid dissociation constant. Which acid is the stronger acid? 1) 0.011 2) 1.52 x 10-2

Calculating Ka from pH – Example #2 A 0.10 M solution of lactic acid (CH3CHOHCOOH) has a pH of 2.44. Calculate the Ka of lactic acid.

Calculating Ka from pH – Whiteboards A student prepared a 0.0075 M solution of HNO2 and measured the pH to be 5.48. Calculate the Ka for HNO2. A 0.020 M solution of niacin (C5H4NCOOH) has a pH of 3.26. What is the acid dissociation constant for niacin? Which acid is the weaker acid? 1. 2. 1.5 x 10-5

Calculating pH from Ka – Example #3 Determine the pH of a 0.30 M solution of CH3COOH, given Ka = 1.8 x 10-5.

Calculating pH from Ka – Example #3 CH3COOH H+ CH3COO- Initial Molarity 0.30 M Change in Molarity -x +x Equilibrium 0.30 – x x Ka = [H+][CH3COO-] = (x)(x) = 1.8 x 10-5 [CH3COOH] 0.30-x The x is so small in comparison to the Molarity of CH3COOH to begin with, we assume Molarity – x is negligible and just drop the x from the formula. (x)(x) = 1.8 x 10-5 0.30

Calculating pH from Ka – Whiteboards Calculate the pH of a 2.92 M HCN solution if the Ka = 6.2 x 10-10. Determine the pH of a 0.095 M hypochlorous acid solution if the acid dissociation constant is 3.5 x 10-8. Which acid is the weaker acid? 1) 4.37 2) 4.24 3) #1