Titration Chapter 19 section 4 Dr. Knorr Honors Chemistry

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

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

Titration The end point is when the indicator changes color. The equivalence point is when the moles of H + = moles of OH -

Strong Acid/Strong Base with Phenolphtalein

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