Titration summary (see figures 19-3 & 19-4) from Monday s lecture:

Transcription

1 Titration summary (see figures 19-3 & 19-4) from Monday s lecture: Strong acid + strong base: 1) initial ph of strong acid calc 2) before equivalence point limiting reactant problem, - ph strong acid calc 3) at equivalence point [acid] = [base] neutral salt no hydrolysis, ph = 7 4) after equivalence point limiting reactant problem, - ph strong base calc big jump in curve, almost flat before and after equivalence point. similar for strong base + weak acid (shape of curve is reversed) weak acid + strong base 1) initial ph of a weak acid (equilibrium calc) 2) before equivalence point limiting reactant problem, - buffer problem (equil. calc.) 3) at equivalence point [acid] = [base] salt in solution but basic salt hydrolysis calc. (equil.) 4) after equivalence point limiting reactant problem, - ph strong base calc. smaller jump in curve, mushy before (buffer zone) equiv. pt. Similar for weak base & strong acid (curve is reversed, fig. 19-5) Usually add strong to weak in these type of titrations Weak acid + weak base smallest jump in ph around equivalence point, not very vertical, difficult to determine visually. Curve is mushy before and afterward due to buffer formation before and after equiv. pt. (figure 19-6).

2 EXAM 3 general topics list Ch identify strong/weak acids & bases - rank relative strength of o weak acids o weak bases o conjugates - KaKb = Kw. Also, pka, pkb. - ph, poh (strong & weak acids & bases). ph + poh = weak acid/base dissociation o writing equations o writing Ka or Kb expressions o calculations - salt hydrolysis o ph of salt solutions o writing equations o writing Ka or Kb expressions o calculations Ch identify buffer - common ion effect - calculate ph of buffer - calculations involving addition of strong acid/base to buffer - preparation of buffers, for example o acid/base addition to weak base/acid o salt addition (especially CAPA #8) - titrations o 4 stages, calculations o understanding of general shape of curves Ch a little. Specifics announced at end of lecture on Monday.

3 Chapter 20 solubility product principle. Ch. 4 review: soluble dissolves in water (solvent) Insoluble doesn t dissolve in water (solvent) General rules (helpful but not necessary to memorize): table 4-8 soluble Na +, K + +, NH 4 - generally soluble F - I - - usually soluble Favorite exception: AgCl SO 2-4, NO - 3, NO - 2, - usually soluble ClO - 3, ClO - 4, CH 3 COO - favorite exception: BaSO 4 Insoluble - some Ag +, Pb 2+, Hg + salts - S 2-, O 2-, OH - many compounds (strong bonding) - Anions with charges greater than 1- (often): o CO 3 2-, PO 4 3-, AsO 4 3-, etc. In equations, represented as (s), espec. when everything else is (aq) AgNO 3 (aq) + NaCl (aq) AgCl (s) + NaNO 3 (aq) In other texts, sometimes AgCl, or AgCl? Mechanistic picture of dissolution of salts (figure from text). Water is polar cations attracted to oxygen part of H 2 O, anions attracted to hydrogen parts of H 2 O. Separates ions. Entropically favored process.

4 Old approach: solubility as on/off switch: either it is, or isn t. Most insoluble things are actually slightly soluble. How soluble, and why, depends on a lot of factors. Soluble salt dissolving: H 2 O NaCl (s) NaCl (aq) Na + (aq) + Cl - (aq) Use these kinds of arrows because process goes to completion, for (strongly) soluble salts. We can also assume that, once dissolved, the ions will separate completely in solution (for the purposes of this chapter.) We usually abbreviate this process to this type of equation: NaCl (s) Na + (aq) + Cl - (aq) For a slightly soluble salt, not all the solid will dissolve, so we can write this as an equilibrium expression. AgCl (s)? Ag + (aq) + Cl - (aq) K = [Ag + ][Cl - ] = K c = K sp sp solubility product. AgCl solid, not in expression. Other examples: Al(OH) 3 (s)? Al 3+ (aq) + 3OH - (aq) K sp = [Al 3+ ][OH - ] 3 coefficients become exponents

5 Bi 2 S 3 (s)? 2Bi 3+ (aq) + 3S 2- (aq) K sp = [Bi 3+ ] 2 [S 2- ] 3 Quiz 5 Write K sp equations for the following slightly soluble compounds: BaSO 4 BaSO 4 (s)? Ba 2+ (aq) + SO 4 2- (aq) K sp = [Ba 2+ ][SO 4 2- ] Ca 3 (PO 4 ) 2 Ca 3 (PO 4 ) 2 (s)? 3Ca 2+ (aq) + 2PO 4 3- (aq) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 Mg(OH) 2 Mg(OH) 2 (s)? Mg 2+ (aq) + 2OH - (aq) K sp = [Mg 2+ ][OH - ] 2 Ag 2 S Ag 2 S (s)? 2Ag + (aq) + S 2- (aq) K sp = [Ag + ] 2 [S 2- ]