In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers.

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1 1 hapter 17 Additional Aspects of Aqueous Equilibria I. Solubility Equilibria The solubility product constant K sp onsider BaSO 4 (s) X Ba 2+ (aq) + SO 4 2- (aq) for which K sp = [Ba 2+ ][SO 4 2- ] K sp is the solubility product. (BaSO 4 is ignored because its concentration is constant.) In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers. Solubility is the amount (grams) of a substance which dissolves to form a saturated solution. Molar solubility is the number of moles of solute dissolving to form a liter of solution. II. Factors that Effect Solubility A.. ommon Ion Effect the solubility of a partially soluble salt is decreased when a common ion is added. onsider the equilibrium established when AgI is added to water AgI(s) XAg + (aq) + I - (aq) K sp = [Ag + ][I-] If a common ion is added, i.e. I - from NaI then [I-] increases and the system is no longer at equilibrium. [Ag + ][I-]> K sp when NaI is added. To return to equilibrium (at the given Ksp) Ag must decrease and therefore the reverse reaction will occur. Solving common ion equilibrium problems follows the same pattern as other equilibrium problems. However, there is an initial concentration of the common ion (from the salt) REMEMBER: Addition of a common ion will cause the equation to shift away from the common ion. B. Solubility and ph Applying Lehatelier s principle: af 2 (s) Xa +2 (aq) + 2F - (aq) If F- is removed, then the equilibrium shifts towards the increase of it and causes solubility of

2 2 caf 2 to increase. How can F- be removed? Addition of a strong acid! F- (aq) + H+(aq) X HF (aq) Therefore, as ph increases [H+] increases and solubility increases. The effect of ph on solubility is tramatic. (Rememeber in labs why we add acids to get things to dissolve!). Formation of omplex ions omplex ions occur under certain conditions, usually metal complexes formed under acid-base interactions. hapter 24 will deal with this extensively. For the time being, complex ions are recognized by a metal ion containing more bonds than it normally would form. The bonds are usually to a polyatomic ion or molecule and carry a charge. [Fe(H 2 O) 5 SN] 2+, [Ag(NH 3 ) 2 ] + The attached NH 3 is called the ligand. (In this case a Lewis base) The fomation of [Ag(NH 3 ) 2 ] + is as follows: Ag+ (aq) + 2NH 3 (aq) X [Ag(NH 3 ) 2 ] + (aq) the equilibrium expression constant is called K f (formation constant) K f = [Ag(NH 3 ) 2 ] + [Ag+][NH 3 ] 2 Effect on solubility: onsider the addition of ammonia to Agl (white precipitate): Agl(s) + NH 3 (aq) XAg(NH 3 ) 3 + (aq) + l-(aq) Effectively, insoluble Agl has been removed from solution. D. Amphoterism Ampotheric oxides will dissolve either a strong acid or a strong base. Examples: hydroxides and oxides of Al 3+, r 3+ Zn 2+, and Sn 2+ the hydroxides general form complex ions with FOUR hydroxide ligands attached to the metal. Al(OH) 3 (s) + OH - (aq) XAl(OH) 4 - (aq) Hydrated metal ions act as weak acids. Thus, amphoterism is interrupted.

3 3 Al(H 2 O) 6 3+ (aq) + OH - (aq) X Al(H 2 O) 5 2+ (aq) + H 2 O(l) Al(H 2 O) 5 2+ (aq) + OH-(aq)X Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O(l) Al(H 2 O) 4 (OH) 2 + (aq) + OH-(aq)X Al(H 2 O) 3 (OH) 3 (s) + H 2 O(l) Al(H 2 O) 3 (OH) 3 (s) + OH-(aq) X Al(H 2 O) 2 (OH) 4 - (aq) + H 2 O(l) III. Buffers omposition and Action of Buffered solutions A buffer consists of a mixture oroughly equal amounts f a weak acid (HX) and its conjugate base (X-) Example: dissolve both acetic acid (H 2 H 3 O 2 ) and sodium acetate (Na+, 2 H 3 O - 2 ) in water. H 2 H 3 O 2 X H H 3 O 2 A buffer has the ability to resist a change in ph when either OH- or H+ is added. I.e. adding 2 H 3 O - 2 increases the concentration of acetate ion in the equilibrium above. reaction shifts to produce more acetic acid until they are in equal ratio again leaving the K a unchanged and therefore ph unchanged. The Buffer capacity and ph Buffer capacity is the amount of acid or base neutralized by the buffer before there is a significant change in ph. Buffer capacity depends on the composition of the buffer. The greater the amounts of the conjugate acid, base pair, the greater the buffer capacity. the ph of the buffer depends on the Ka. If ka is sufficiently small ( the equilibrium concentration of the undissociated acid is close to the initial concentration), then K a = [H+][A-] [HA] So, [H+] =( K a )[HA]/[A-] Taking negative logs we get -log [H+] = -logk a - log [HA]/[A-] ph = pk a -log [HA]/[A-] Or ph pk a +log [base]/[acid] Henderson-Hasselbalch equation Addition of Strong acids or Bases to Buffers We break the calculation into two parts: stoichiometric and equilibrium The amount of strong acid or base added results in a neutralization reaction: X- + H 3 O + 6HX + H 2 O HX + OH- 6 X- + H 2 O By knowing how much H 3 O+ or OH- was added (stoichiometry) we know how much HX or X- is formed. With the concentrations of HX and X- (note the change in volume of solution) we can calculate ph from the Henderson hasselbalch equation.

4 4 Example 1: What is the ph of a buffer that is 0.12 M in lactic acid, H 3 H 5 O 3, and 0.10 M in sodium lactate? For lactic acid, K a = 1.4 x solution: pk a = -log 10 K a ph = pk a + log ([base]/[acid]) = log (0.10/0.12) ph = 3.77 Example 2: A buffer is made by adding mol H 2 H 3 O 2 and M Na 2 H 3 O 2 to enough water to make 1 liter solution. The ph of the buffer is alculate the ph of this solution (a) after mol of NaOH is added (neglect any volume changes) Solution (a) Step 1: Stoichiometric calculation: The OH- provided by NaOH reacts with H 2 H 3 O 2. The following table summarizes the concentrations before and after neutralization: H 2 H 3 O 2 (aq) + OH - H 2 O (l) + 2 H 3 O - 2 (aq) Before rxn M M M hange M M M After Rxn 0.280M 0.0M M Step 2: Equilibrium calculation: the solution after neutralization contains H 2 H 3 O 2-2 H 3 O 2 conjugate acid-base pair. We next consider the proton-transfer equilibrium in order to determine the ph of solution. H 2 H 3 O 2 (aq) + H 2 O H 3 O + (l) + 2 H 3 O - 2 (aq) Before rxn M M M hange - x M x M + x M After Rxn ( x) M x M ( x) M K a = [H 3 O + ][ 2 H 3 O 2 - ] / [H 2 H 3 O 2 ] = (x) ( x)/( x) = (x) (0.320)/0.280 = 1.8 X 10-5 * x = [H 3 O + ] = {(0.280)(1.8 x 10-5 )}/0.320 = 1.6 x 10-5 M

5 5 ph = -log (1.6 x 10-5 ) = 4.80 *In step 2, you could also use the Henderson-Hasselbalch equation: ph = pk a + log [base]/[acid] pk a = -log (1.8 x 10-5 ) = log [0.320]/[0.280] = 4.80 IV. Precipitation and Separation of ions onsider BaSO 4 X Ba 2+ + SO 4 2- At any instant in time Q = [Ba 2+ ][SO 4 2- ] 1. If Q<K sp precipitation occurs until Q = K sp 2. If Q = K sp, equilibrium exists 3. If Q > K sp, solid dissolves until Q=K sp Based on solubilities, ions can be selectively removed from solutions. onsider a mixture of Zn 2+ 9aq) and u 2+ (aq). us (K sp = 6 x ) is less soluble than ZnS (K sp = 2 x ). us will be removed from solution before ZnS. As H 2 S is added to the green solution, black us forms in a colorless solution of Zn 2+. When more H 2 S is added, a second precipitate of white ZnS forms Selective precipitation of Ions Ions can be separated from each other based on their salt solubilities. Example: if Hcl is added to a solution containing u 2+ and Ag +, the silver precipitates (K sp = 1.8 x ) while the u 2+ remains in solution.>>>>>selective Precipitation V. Qualitative analysis remember old lab: review how you detect certain ions Qualitative analysis :designed to detect the presence of metal ions Quantitative analysis: determine how much metal is present. VI. Acid-Base titrations Titration curve: a plot of ph vs volume during a titration (we will be doing this in lab) A. Strong acid-strong base At equivalence point, ph = 7 Indicator: Phenolphthalein changes color between 8.3 and 10.0

6 6 In acid colorless, in base pink Equivalence point : the point at which the acid and base are present in stoichiometric mole relationship Endpoint: the observed end of the titration (color change) Titration error: the difference between the observed and equivalence point. B. Strong base, weak acid acetic acid and sodium hydroxide leaves water and acetate ion which is basic. Therefore the titration ends with a basic ph greater than 7.. Weak acid- Weak Base titration At the equivalence point, there is a steep change in ph and then a leveling off due to buffer effects. For very weak acids it is impossible to detect the equivalence point. D. Polyprotic acids In a titration there are n equivalence points based on the number of ionizable H