1. Calculating pH and pOH for WEAK acids and bases Just when you were thinking things were easy….

2. Just Listen for a few minutes  Remember that Strong acids and Bases are different from Weak Acids.  Every bit of STRONG acids or bases put into water completely split into the [H+] or [OH-] ions.  However for WEAK acids and weak bases, we will need to determine how much of the ions dissociate in water.

3. Equilibrium & K  A weak substance will only partially dissociate and will do so until the reaction reaches equilibrium.  We will solve for the K a for acids and K b for bases. (Dissociation Constants)  These values represent the relationship between the concentrations of the products over the reactants when equilibrium is reached.

4. A quick reminder  How we solve for K is the law of mass action: Ka = [Y] c [Z] d [W] a [X] b  The brackets[ ] represent the molar concentrations. aW + bX cY + dZ

5. A K A Equilibrium Expression:  Were going to write the “Blueprint” equation for any acid dissociation:  HA  H + + A -  Any acid  H + ion + Conj. Base  What would the K A = ?  K A = [H + ][A - ] [HA]

6. Reaction Practice:  Lets write a weak acid reaction and practice:  First lets predict the products:  HCOOH + H 2 O ↔ H + + HCOO -  * You need to remember from equilibrium calculations, that pure liquids are not included.  HCOOH ↔ H + + HCOO -  Notice the arrow goes both ways.

7. Equilibrium Expression  Lets write an EQ. expression. What will K A = ?  HCOOH ↔ H + + HCOO -  K A = [H + ][HCOO - ] [HCOOH]

8. Ice Ice Baby  The tool for solving equilibrium problems is the ICE chart.  It’s broken into three sections: I = Initial concentrations C = Change in concentrations E = Equilibrium concentrations  Each of these is recorded in Molarity or concentration of the solution.

9. Ice Chart Template: AcidH + IonConjugate Base Initial Change Equilibrium

10. Solving for the K A  Solving for K A can be done if you know the pH or [H + ] of the solution at equilibrium.  A student prepares a solution from 0.10 M formic acid (HCOOH) and found its pH at equilibrium to be 2.38. Determine the K A.  To solve for the K A, we need first need the equilibrium expression and then we can use an ICE table.  HCOOH ↔ H + + HCOO -

11. Ice Chart: HCOOHH+H+ HCOO - Initial Change Equilibrium

12. Initially:  A student prepares a solution from 0.10 M formic acid (HCOOH) and found its pH at equilibrium to be 2.38. Determine the K A.  Initially we know the concentration of the formic acid = 0.10 M.  But think about it, AT THE VERY BEGINNING, how much of the products have we made?  None! So we complete the Initial Conc.

13. Ice Chart: HCOOHH+H+ HCOO - Initial 0.10 M0.0 M Change Equilibrium

14. Change in Concentrations  We don’t know the numbers to this part yet, but we do know something about it.  When this acid molecule dissociates, it will create one hydrogen ion, and one molecule of the conjugate base.  Not always the case, but true for this acid!  HCOOH ↔ H + + HCOO -  We put X in for the change in amounts: - X for decreasing amounts (HCOOH) +X for increasing amounts. (H + ) and (HCOO - )

15. Ice Chart: HCOOHH+H+ HCOO - Initial 0.10 M0.0 M Change - X+ X Equilibrium

16. Equilibrium Concentrations  This is pretty easy. Read straight down the column.  Keep the same signs.

17. Ice Chart: HCOOHH+H+ HCOO - Initial 0.10 M0.0 M Change - X+ X Equilibrium0.10 - XXX

18. Final Concentrations  But we know something about the final concentration!  A student prepares a solution from 0.10 M formic acid (HCOOH) and found its pH at equilibrium to be 2.38. Determine the K A.  pH can tell us [H + ] so we can use that to solve for X in the ICE table and plug it in!  pH = -log[H+]  10 -2.38 = 0.0042657 *Dolphins in San Fran  4.3 x 10 -3 [H + ] * This is the [H + ] and X!!!

19. Ice Chart: HCOOHH+H+ HCOO - Initial 0.10 M0.0 M Change - 4.3 x10 -3 4.3 x10 -3 Final 0.10 – 4.3 x10 -3 4.3 x10 -3

20. Finally: solving for the K A  Using the equation we need to write the equilibrium expression: We will plug in concentrations from the “Final” part of the Ice Chart:  K A = [H + ][HCOO - ] [HCOOH]  K A = [4.3 x10 -3 ][4.3 x10 -3 ] [0.10 – 4.3 x10 -3 ]  K A = 1.9 x 10 -4