Chapter 16: Applications of Aqueous Equilibrium Part 2. Acid-Base Titrations

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1 Chapter 16: Applications of Aqueous Equilibrium Part 2 Acid-Base Titrations When you add an acid and a base together, a neutralization rxn occurs. In the lab, we do neutralization rxns all the time as titrations. In a titration, a known volume of an acid (or base) is added to a flask, and then a known concentration of a base (or acid) is added slowly from a buret (titrated). The titration is followed with a ph meter and/or an acid/base color indicator so that it is known when the neutralization is complete (UV-Vis, or voltage may also be followed). The point at which the acid and base have neutralized each other stoichiometrically is called the equivalence point. If an acid/base color indicator is used, there is a color change (and ph change) when the endpoint is reached, which should be very close to the equivalence point (if you picked the right indicator). The half-equivalence point is when the concentrations of the acid and its conjugate base are equal to each other, and this is where the ph equals the pka. Titrations are useful for several purposes: To determine the K a or K b of an acid or base To determine the molecular weight of an acid or base To standardize, or determine accurately, the concentration of an acid or base. There are 4 types of titrations commonly done in the lab: Strong acid-strong base Weak acid-strong base Weak base-strong acid Polyprotic acid-strong base Strong Acid-Strong Base NaOH + HCl NaCl + H 2 O Typically, an acid of unknown concentration is titrated with a base of known concentration (although the reverse is also done). As it is a 1:1 mole ratio between the base and the acid, it is easy to calculate how many moles of acid were neutralized with the base. As we titrate, we measure the amount of base added in ml and the ph. Thus, a titration curve is obtained: What do you notice about the curve? There are 4 regions of the curve: 1) before adding OH - so starting [acid] = H + ; 2) before the equivalence point, so there is excess H + ; 3) at the equivalence point where the H + and OH - have neutralized each other; and 4) after the equivalence point so there is excess OH -

2 Typical problem: Calculate the ph of ml of a M nitric acid sln 1) before titration with KOH; 2) after ml of 0.100M KOH is added; 3) after ml of the KOH sln has been added; and 4) after ml of the KOH sln has been added. Weak Acid-Strong Base HF + KOH KF + H 2 O Here is a typical titration curve for this common titration: How is this curve different from the titration curve for the strong acid/strong base titration?

3 Typical problem: Calculate the ph of ml of a M benzoic acid sln 1) before titration with KOH; 2) after ml of 0.100M KOH is added; 3) after ml the KOH sln has been added; and 4) after ml of the KOH sln has been added. pk a = 4.20 Another typical problem: 3.467g of an unknown monoprotic acid was diluted to ml with water. It was then titrated with M NaOH. After ml of NaOH was added, the ph was 5.25; and the equivalence point was reached after ml was added. Calculate the molecular weight of the unknown acid; and calculate its pk a.

4 The weaker the acid, the shallower the curve, and the more difficult it is to determine the correct pk a accurately. Weak Base-Strong Acid HCl + NH 3 NH 4 Cl This titration is the opposite of the last, so the calculations are similar. Draw the titration curve for a general weak base-strong acid titration (base in flask, acid in buret) Problem: What is the ph of ml of a M weak base sln 1) before titration begins; 2) after ml of M HCl is added; 3) after ml of M HCl is added; 4) after ml of the HCl is added; and 5) after ml of the HCl is added. pk b of the weak base = Note: part 2 and 3 are both in the buffer region, but part 2 is much easier than part 3. Why?

5 Polyprotic Weak Acid Titrations with Strong Base H 2 SO 3 + 2NaOH 2 H 2 O + Na 2 SO 3 Polyprotic acids have more than 1 acidic protons and so have a more complicated titration curve: What do you notice about this titration curve? What is interesting for a diprotic acid is that at the 1 st equivalence point, HA - is at a maximum while H 2 A and A 2- are equal. So: ph = pk a1 + pk a2 2