Aqueous Equilibria, Part 2 AP Chemistry Lecture Outline

Transcription

1 Aqueous Equilibria, Part 2 AP Chemistry Lecture Outline Name: The Common-Ion Effect Suppose we have a weak acid and a soluble salt of that acid. CH 3 COOH NaCH 3 COO CH 3 COOH CH 3 COO + H + Since NaCH 3 COO is strong, adding it to the solution By Le Chatelier The result is that This illustrates the common-ion effect: The dissociation of a weak electrolyte decreases when a strong electrolyte that has an ion in common with the weak electrolyte is added to the solution. EX. Find the ph of a solution containing M HNO 2 (K a = 4.5 x 10 4 ) and 0.10 M KNO 2. Buffered Solutions ( buffers ) contain a weak conjugate acid-base pair, i.e., -- contain an acidic species and a basic species that do NOT consume each other through neutralization e.g., For a CH 3 COOH-and-CH 3 COO buffer, you could use For an NH + 4 -and-nh 3 buffer, you could use 1

2 buffer capacity: the amount of acid or base the buffer can neutralize before the ph begins -- to change appreciably EX. Find the ph of a buffer that is 0.12 M lactic acid, HC 3 H 5 O 3 (K a = 1.4 x 10 4 ) and 0.10 M sodium lactate. Alternately, you could use the Henderson-Hasselbalch equation: ph pka log base acid -- Addition of Strong Acids or Bases to Buffers -- Reactions between strong acids/bases and weak bases/acids proceed to completion. -- We assume the strong acid/base is completely consumed. -- When adding a strong to buffered solutions (1) (2) 2

3 EX. A buffered solution of ph 4.74 contains 0.30 mol CH 3 COOH (K a = 1.8 x 10 5 ) and 0.30 mol NaCH 3 COO. Calculate the ph after mol NaOH is added. Ignore volume changes. Acid-Base Titrations equivalence point: Strong Acid Strong Base Titrations Any indicator whose color change begins and ends along the vertical line is okay. -- phenolphthalein (ph ) base = acid = -- methyl red (ph ) base = acid = ph 7 ph curve for HCl titrated with NaOH ml of NaOH added 3

4 EX. Find ph when ml of 0.10 M HNO 3 are mixed with ml of 0.10 M KOH. Weak Acid Strong Base Titrations The equivalence point is when, say, 50.0 ml of 0.10 M NaOH have been added to 50.0 ml of 0.10 M CH 3 COOH, but ph is > 7 at that point because... ph 7 ph curve for CH 3 COOH titrated with NaOH ml of NaOH added ph curve for H 2 CO 3 Titration curves for polyprotic acids (e.g., H 2 CO 3 ) look something like -- they have... ph 4 ml of base added

5 EX. Calculate the ph when 10.0 ml of M NaOH are added to 40.0 ml of M benzoic acid (C 6 H 5 COOH, K a = 6.3 x 10 5 ). EX. Calculate the ph at the equivalence point when 40.0 ml of M C 6 H 5 COOH (K a = 6.3 x 10 5 ) are titrated with M NaOH. 5

6 Solubility Equilibria -- involve the dissolution or precipitation of ionic compounds Consider a saturated solution of barium sulfate: For this case, the solubility-product constant is equal to: -- K sp is the equilibrium constant between undissolved and dissolved ionic solute in a saturated aqueous solution EX. Write the solubility-product constant expression for calcium fluoride. EX. Copper (II) azide has K sp = 6.3 x Find the solubility of Cu(N 3 ) 2 in water, in g/l. 6

7 Factors That Affect Solubility 1. For solids, as temperature increases, solubility common-ion effect Use Le Chatelier s principle. For example, with... CaF 2 (s) Ca 2+ (aq) + 2 F (aq) 3. ph and solubility Compounds with anions exhibiting basic properties (e.g., Mg(OH) 2 / OH, CaCO 3 / CO 2 3, CaF 2 / F ) in solubility as solution becomes more acidic. 4. presence of complex ions: metal ions and the Lewis bases bonded to them e.g., AgCl(s) + 2 NH 3 (aq) Ag(NH 3 ) + 2 (aq) + Cl (aq) In general, the solubility of metal salts in the presence of suitable Lewis bases (e.g., NH 3, CN, OH ) if the metal forms a complex ion with the bases. 5. amphoterism Many metal hydroxides and oxides are amphoteric. They are insoluble at ph ~ 7, but will dissolve in strongly acidic or strongly basic solutions. (Recall that many metal ions act like acids in solution.) As an example... Al(H 2 O) 3+ 6 (aq) + OH (aq) Al(H 2 O) 5 (OH) 2+ (aq) + H 2 O(l) If we continue adding more and more OH (aq), we ll get... Al(H 2 O) 5 (OH) 2+ (aq) + OH (aq) Al(H 2 O) 4 (OH) + 2 (aq) + H 2 O(l) Al(H 2 O) 4 (OH) + 2 (aq) + OH (aq) Al(H 2 O) 3 (OH) 3 (s) + H 2 O(l) Al(H 2 O) 3 (OH) 3 (s) + OH (aq) Al(H 2 O) 2 (OH) 4 (aq) + H 2 O(l) Al(H 2 O) 2 (OH) 4 (aq) + OH (aq) Al(H 2 O)(OH) 2 5 (aq) + H 2 O(l) Al(H 2 O)(OH) 2 5 (aq) + OH (aq) Al(OH) 3 6 (aq) + H 2 O(l) Often, the H 2 O is left out. 7

8 Precipitation and Separation of Ions For BaSO 4 (s) Ba 2+ (aq) + SO 4 2 (aq)... K sp = [ Ba 2+ ] [SO 4 2 ] We could reach equilibrium from the left... or from the right... At any given time, the ion product Q = [ Ba 2+ ] [SO 2 4 ] If Q > K sp... If Q < K sp... If Q = K sp... selective precipitation: using the different solubilities of ions to separate them Consider a solution with Ag + and Cu Add HCl. The K sp of AgCl = 1.8 x 10 10, but CuCl 2 is soluble (i.e., Cu 2+ doesn t precipitate). -- Then... EX. Will a precipitate form from mixing 0.10 L of 8.0 x 10 3 M Pb(NO 3 ) 2 (aq) and 0.40 L of 5.0 x 10 3 M Na 2 SO 4 (aq)? 8

9 Qualitative Analysis for Metallic Elements Steps: (1) (2) aqueous solution containing god-knowswhat metal ions Add HCl. insoluble chlorides (AgCl, Hg 2 Cl 2, PbCl 2 ) ppt out. Filter and test for a specific ion. Bubble H 2 S(g) in. acid-insoluble sulfides (CuS, Bi 2 S 3, CdS, PbS, HgS, As 2 S 3, Sb 2 S 3, SnS 2 ) ppt out. Filter and test for a specific ion. Add (NH 4 ) 2 S. base-insoluble sulfides and hydroxides (Al(OH) 3, Fe(OH) 3, Cr(OH) 3, ZnS, NiS, CoS, MnS) ppt out. Filter and test for a specific ion. Add (NH 4 ) 2 PO 4. insoluble phosphates (Ba 3 (PO 4 ) 2, Ca 3 (PO 4 ) 2, MgNH 4 PO 4 ) ppt out. Filter and test for a specific ion. 9