Titration Chapter 19 section 4

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Presentation transcript:

Titration Chapter 19 section 4 Honors Chemistry

Neutralization Strong acid + strong base = salt +H2O HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l) 2HNO3 (aq) + Ca(OH)2  Ca(NO3)2 (aq) + 2H2O (l) How many moles of potassium hydroxide are needed to completely neutralize 1.56 moles of phosphoric acid? H3PO4 (aq) + 3 KOH (aq)  K3PO4 (aq) + 3H2O (l) 1.56 moles H3PO4 x 3moles KOH/1mole H3PO4 = 4.68 moles KOH

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

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

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

Calculating Concentration How many milliliters of 0.45M HCl will neutralize 25.0ml of 1.00M KOH? HCl (aq) + KOH (aq)  KCl (aq) + H2O (l) 25.0ml KOH x 1mol KOH x 1molHCl x 1000mL 1000ml 1molKOH 0.45mole HCl = 56 mL HCl