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See summary: top of p.778 in textbook

Half-Equivalence Point Volume of titrant added is half the volume of the equivalence point pH = pKa Why? pH = pKa + log (CB / A) At the half equivalence point, there is the same amount of original acid and conjugate base (formed from neutralization).

Half-Equivalence Point Volume of titrant added is half the volume of the equivalence point Moles of strong base added is half that of original weak acid HA + OH-  A- + H2O .50 .25 -.25 -.25 +.25 .25 .25 pH = pKa + log .25 pH = pKa @ half-equivalence point

Base Titrated with Strong Acid

Weak Base Titrated with Strong Acid Initial pH: use Kb for weak base, ICE table Between initial and equiv pt: NH3 + H3O+  NH4+ + H2O Complete reaction stoichiometry. Use in H-H equation.

Weak Base Titrated with Strong Acid 3. At equiv pt: All NH3 reacted. NH4+ has been formed. NH4+ + H2O  NH3 + H3O+ Use Ka and ICE table 4. After equiv pt: Any additional H+ added has nothing to react with. Use excess H+ to calc pH

Acid-Base Titrations Add solution (either the acid or the base) from buret to solution in flask The concentration of one solution is known. One concentration is unknown. Often, the goal is to find the unknown concentration.

What is the concentration of NaOH when 30 What is the concentration of NaOH when 30.0mL of the base neutralizes 26mL of 0.50M HCl? When 25 mL of HCl is titrated with 6.00 M Ba(OH)2, it takes 39.3 mL of base to reach the end point. What is the pH of the acid solution?

What is the concentration of NaOH when 30 What is the concentration of NaOH when 30.0mL of the base neutralizes 26mL of 0.50M HCl?

When 25 mL of HCl is titrated with 6. 00 M Ba(OH)2, it takes 39 When 25 mL of HCl is titrated with 6.00 M Ba(OH)2, it takes 39.3 mL of base to reach the end point. What is the pH of the acid solution?