Acid Base Titrations Chm 3.2.3.

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Presentation transcript:

Acid Base Titrations Chm 3.2.3

Titration The controlled addition and measurement of the amount of a solution with a known concentration required to react completely with a measured amount of a solution with an unknown concentration

H+ + Cl- + Na+ + OH-  H2O + Na+ + Cl- Acid Base Reaction HCl + NaOH  H2O + NaCl H+ + Cl- + Na+ + OH-  H2O + Na+ + Cl- H+ + OH-  H2O In an acid base neutralization reaction H+ in being neutralized by OH- to form H2O

Acid Base Stoichiomety What concentration of 100 mL of an NaOH is needed to neutralize 200 mL of a 0.5 M HCl solution. 200 mL of 0.5 M HCl = 0.10 mole HCl = 0.1 moles H+ H+ + OH-  H2O 0.1 moles OH- = 0.1 mole NaOH in 100 mL = 1.0 M solution of NaOH

When H+ and OH- are 1:1 mole ratio For a neutralization reaction MAVA = MBVB Proven by previous example: (0.5 M HCl x 200 mL) = (1.0 M NaOH x 100 mL)

Example What volume of a 0.3 M HCl solution will be needed to neutralize 500 mL of a 0.1 M KOH solution?