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TITRATION OF A WEAK ACID WITH A STRONG BASE

Mistakes are a part of being human. Appreciate your mistakes for what they are: precious life lessons that can only be learned the hard way. Unless it's a fatal mistake, which, at least, others can learn from.        - Al Franken -. TITRATION OF A WEAK ACID WITH A STRONG BASE

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TITRATION OF A WEAK ACID WITH A STRONG BASE

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  1. Mistakes are a part of being human. Appreciate your mistakes for what they are: precious life lessons that can only be learned the hard way. Unless it's a fatal mistake, which, at least, others can learn from.       - Al Franken -

  2. TITRATION OF A WEAK ACID WITH A STRONG BASE TITRATIONS OF WEAK ACIDS WITH A STRONG BASE OR WEAK BASES WITH A STRONG ACID GET A LITTLE MORE COMPLICATED AS ONE MUST CONSIDER THE EQUILIBRIUM OF THE WEAK ACID OR WEAK BASE. THE CALCULATIONS LEADING UP TO THE EQUIVALENCE POINT WOULD BE SIMILAR TO THE CALCULATIIONS WE MADE FOR BUFFER SOLUTIONS. THE MOST IMPORTANT THING WITH THESE TITRATIONS WILL BE THE pH AT THE EQUIVALENCE POINT, AS THIS WILL DETERMINE WHAT INDICATOR WE WOULD SELECT FOR THE TITRATION OR WHAT pH WE WOULD BE CONCERNED WITH ON OUR pH METER.

  3. LET’S CONSIDER THE TITRATION OF 40 ml OF 0.2 M ACETIC ACID, HC2H3O2, WITH 0.1 M NaOH. AT THE EQUIVALENCE POINT, Va x Ma = Vb x Mb SO 40 X 0.2 = 0.1 X Vb and Vb = 8/0.1 = 80 ml AT THE EQUIVALENCE POINT, YOUR MAJOR SPECIES ARE C2H3O2- (Ac- ion), Na+, and H2O THE EQUILIBRIUM WE WILL BE CONCERNED WITH IS Ac- + H2O == Hac + OH- Kb = [Hac][OH-]/[Ac-] WE COULD SHOW THAT KB = KW/Ka = 10-14/(1.8 x 10-5) Kb = 5.56 x 10-10 = [Hac][[OH-]/[Ac-]

  4. BEFORE EQUILIBRIUM [Ac-] = 8 mmole/120 ml = 6.67 x 10-2 M AT EQUILIBRIUM [Hac] = [OH-] = x and [Ac-] = (6.67 x 10-2) – x Kb = x2/(6.67 x 10-2 – x) and if 6.67 x 10-2 >> x Kb = x2/6.67 x 10-2 = 5.56 x 10-10 so x2 = (5.56 x 10-10)(6.67 x 10-2) X2 = 3.77 x 10-11 and x = [OH-] = 6.14 x 10-6 pOH = -log(6.14 x 10-6) = 5.21 pH = 14 – 5.21 = 8.79 WHY IS THE EQUIVALENCE POINT FOR THIS TITRATION NOT AT 7, WHICH WOULD BE THE pH OF A NEUTRAL SOLUTION (PURE WATER)??

  5. YOU ARE DEALING WITH THE SALT OF A WEAK ACID. THE ANION OF A WEAK ACID IS ITS CONJUGATE BASE. IT HYDROLYZES TO GIVE A SLIGHTLY BASIC SOLUTION.

  6. IF YOU WERE SELECTING AN INDICATOR, WHICH ONE MIGHT YOU SELECT?

  7. THE SITUATION WOULD BE SIMILAR IF YOU WERE TITRATING A WEAK BASE WITH A STRONG ACID. FOR EXAMPLE, IF YOU WERE TITRATING AMMONIA, NH3, WITH HCl. NH3 + HCl  NH4+ + Cl- THE MAJOR SPECIES AT THE EQUIVALENCE POINT WOULD BE NH4+, Cl-, H2O NH4+ IS THE CONJUGATE ACID OF THE WEAK BASE NH3, SO IT WOULD UNDERGO HYDROLYSIS NH4+  NH3 + H+ KA = [NH3][H+]/[NH4+] = Kw/Kb WE SOULD EXPECT THE EQUIVALENCE POINT TO BE SLIGHTLY ON THE ACIDIC SIDE (LESS THAN 7).

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