NaOH  OH- + Na+ EXAM #1 M = mol #L 0.10 = mol 0.008 M = mol #L EXAM #1a: Find moles of OH- in 8.00mL of 0.10M NaOH. NaOH  OH- + Na+ M = mol #L 0.10 = mol 0.008 Mol (OH-) = 8 * 10-4 EXAM #1b: Find molarity of OH- AFTER of 8.00mL of 0.10M NaOH is titrated in. NOTE total volume = Vacid(50mL) + Vbase(8 mL) = 58 mL M = mol #L M = 8*10-4mol .058L M = 0.0138

0.005 mol 8.0*10-4mol -X X 0.005-X 8*10-4 M = mol #L 0.10 = mol 0.050L EXAM #1c: Find the pH after 8.00 mL of 0.10 M OH- is added to the 50.0 mL acid sample. FIRST FIND MOLES OF ACID ORIGINALLY PRESENT. THEN EVALUATE NEUTRALIZATION STOICHIOMETRY; REMEMBER THE OH- IS LIMITING UP TO EQ POINT. Working in moles alleviates the need to keep track of dilutions! M = mol #L 0.10 = mol 0.050L Mol HA= 0.005 HA OH-  A- H2O 0.005 mol 8.0*10-4mol -X X 0.005-X 8*10-4 NOW USE THE FINAL MOLES IN HENDERSON-HASSLEBACH OH- IS LIMITING REAGENT AND IS CONSUMED TOTALLY

EXAM#1-c CONTINUED; HA  A- H+ 0.005 8.0*10-4 -X X 0.005-X 8.0*10-4 +X pH = pKa + LOG [A-] [HA] pH = 9.21 + LOG [8.0*10-4 ] [0.0042] Ok to work in moles for this equation. pH = 8.49

EXAM #1d and e: AT THE pKa (HALFWAY) POINT: Moles OH- added =: mole HA neutralized moles HA remaining A- produced *********AND mol OH- added = ½ HA INITIAL ******** HA OH-  A- H2O 0.005 mol 0.0025 -X X 0.005-X=0.0025 Neutralization stoichiometry: base added limits, OH- added = X and is completely consumed #1e) Moles OH- at halfway point = ½ HA original ½ of 0.005 = 0.0025

pH = pKa + LOG [A-] [HA] pH = 9.21 + LOG [0.0025] [0.0025] #1-f) AT THE pKa (HALFWAY POINT) point pH = pKa pH = pKa + LOG [A-] [HA] pH = 9.21 + LOG [0.0025] [0.0025] pH = pKa = 9.21 #1-g) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0.0 MOLE, AND A- IS THE ONLY ION REMINING. OH- added total) = HA original = A- produced

0.005 mol 0.0050 -X X 0.005 HA OH-  A- H2O M = mol #L #1-h) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0.0 MOLE, AND A- IS THE ONLY ION REMINING. OH- added (total) = HA original = A- produced = 0.005 mol HA OH-  A- H2O 0.005 mol 0.0050 -X X 0.005 M = mol #L 0.10 = 0.005 mol X L vol OH- = .050 L = 50.0mL

#1-g) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0 #1-g) AT THE EQUIVALENCE POINT HA IS TOTALLY NEUTRALIZED TO 0.0 MOLE, AND A- IS THE ONLY ION REMINING. OH- added total) = HA original = A- produced, The Ph at equivalence is calculated by the hydrolysis equilibrium of A-. THE EQUIVALENCE POINT STOICHIOMETRY GIVES THE A- MOLARITY FOR THE EQUIVALENCE (HYDROLYSIS OF A-) EQUILIBRIUM. HA OH-  A- H2O 0.005 mol 0.0050 -X X 0.005 HYDROLYSIS EQUILIBRIUM OF A- AT EQUIVALENCE PT. A- HOH  HA OH- 0.005 mol IGNORE SOLVENT -X X 0.005-X

HYDROLYSIS EQUILIBRIUM OF A- AT EQUIVALENCE POINT A- HOH  HA OH- 0.005 mol IGNORE SOLVENT -X X 0.005-X Kb = Kw/Ka = [HA][OH-] [A-] Kb = 1.6 x 10-5 = (X)(X) ; X = 8.86 x 10-4 = [OH-] (0.005/0.1L) MUST CONVERT MOLES TO MOLARITY, TOTAL VOLUME IS 0.10 L (0.05L ACID AND 0.05L BASE) pOH- = -LOG(8.86 x 10-4) = 3.05 pH = 10.95 at equivalence