1. Titration Chapter 19 section 4 Dr. Knorr Honors Chemistry

2. Neutralization Strong acid + strong base = salt +H 2 O – HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) – 2HNO 3 (aq) + Ca(OH) 2  Ca(NO 3 ) 2 (aq) + 2H 2 O (l) How many moles of potassium hydroxide are needed to completely neutralize 1.56 moles of phosphoric acid? H 3 PO 4 (aq) + 3 KOH (aq)  K 3 PO 4 (aq) + 3H 2 O (l) 1.56 moles H 3 PO 4 x 3moles KOH/1mole H 3 PO 4 = 4.68 moles KOH

3. Titration A solution of unknown concentration (analyte) is placed in a flask. A solution of known concentration (titrant) is placed in buret. Titrant is added to analyte until indicator changes color http://water.me.vccs.edu/courses/env211/changes/titration.gif

4. Titration The end point is when the indicator changes color. The equivalence point is when the moles of H + = moles of OH - http://img.sparknotes.com/figures/3/3a5994498f24d59f5d5d762b40844a2a/sasb.gif

5. Strong Acid/Strong Base with Phenolphtalein http://2.bp.blogspot.com/_Red3kx4ddLw/THXg7KPCCKI/AAAAAAAAANo/DyCNzfO0Q0c/s1600/Titration.gif

6. Calculating Concentration How many milliliters of 0.45M HCl will neutralize 25.0ml of 1.00M KOH? HCl (aq) + KOH (aq)  KCl (aq) + H 2 O (l) 25.0ml KOH x 1mol KOH x 1molHCl x 1000mL 1000ml 1molKOH 0.45mole HCl = 56 mL HCl