MRA - MOTIVATION (continued) [pages 6 & 7 of Acid-Base Handout]

Transcription

1 SOLVING ACID-BASE EQUILIBRIA: The MAIN REACTION APPROXIMATION (MRA) [page 6 of Acid-Base Handout] MOTIVATION: Aqueous solutions containing more than pure water will have - at least - 2 equilibria occurring SIMULTANEOUSLY. To be rigorous, equilibria must be solved SIMULTANEOUSLY [N-equations, N unknown equilibrium concentrations]. [See Sec. 7.9, text.] This can be done MRA - MOTIVATION (continued) Although this can be done, this approach often becomes excessively numerical and - more importantly, CHEMICAL INTUITION is lost. We can take advantage of the fact that often, these multiple equilibria had K eq values that are far apart, i.e., 10 2 or greater K eq,1 100 K eq,2 100 K eq,3 This correlates with another observation 1 2 MRA - MOTIVATION (continued) Important reactions can only occur between species that are in APPRECIABLE INITIAL concentration. Thus, If we can identify the INITIAL SPECIES that are appreciably present when an aqueous solution is prepared - and CLASSIFY their behavior (ACID, BASE, ), we can MRA - MOTIVATION (continued) Construct ALL possible reactions (for now, Brønsted-Lowry (B-L) acid-base rxns) & Determine the K eq value of each competing rxn. [Recall how to use K a s, K b s, & K W via Hess.] The rxn with the LARGEST K eq value is termed the MAIN REACTION and as such dominates all other equilibria of smaller K eq values. 3 4

2 MRA - MOTIVATION (continued) After the Main Reaction equilibrium is solved A SEQUENTIAL procedure is followed, i.e., The rxn with next largest K eq value is solved subject to the MRA prescription that Any species already equilibrated in a rxn of larger K eq value - does NOT CHANGE when appearing in any subsequent rxn of smaller K eq value. This simplifies the problem considerably as we progress down the chain. MRA - Picking INITIAL SPECIES [page 7 of Acid-Base Handout] Before we do an example or 2, we must see how to determine the INITIAL SPECIES via the MRA. For now, the possible classifications are: Aqueous Acids (according to B-L). Aqueous Bases (according to B-L). Aqueous Salts (Ionic Compounds - already dissolved. We will deal with un-dissolved (solid) ionic compounds when we cover solubility in Chapter 8.) MORE 5 6 Aqueous STRONG Acids (K a ~ 10 +x ): - completely dissociate as initial species. Some common strong acids. LEARN THESE!: HALO-ACIDS HX(aq) [X = Cl -, Br -, I -, NOT F - ] HX(aq) --(completely)--> H + (aq) + X - (aq). HNO 3 (aq) --(completely)--> H + (aq) + NO 3- (aq). HClO 4 (aq) --(completely)--> H + (aq) + ClO 4- (aq). H 2 SO 4 (aq) --(PARTIALLY)--> H + (aq) + HSO 4- (aq). Aqueous STRONG BASES - OH - bases (K b ~ 10 +x ): - completely dissociate as initial species. Some common strong bases. LEARN THESE!: Group 1A Metal Hydroxides (Except H 2 O) [MOH(aq) with M = Li, Na, K, Rb, Cs, Fr] MOH(aq) --(completely)--> M + (aq) + OH - (aq). Group 2A Metal Hydroxides (Except Be(OH) 2 ) [M(OH) 2 (aq) with M = Mg, Ca, Sr, Ba, Ra] M(OH) 2 (aq) --(completely)--> M +2 (aq) + 2 OH - (aq). 7 8

3 Aqueous Weak Acids (K a ~ 10 -x ): - remain UNDISSOCIATED as initial species. An acid that isn t strong is presumed weak (can always check the K a value ~ 10 -x ). Some Examples ( undissociated initially ): HF(aq) (K a = 7.2 x 10-4 ). CH 3 COOH(aq) (K a = 1.8 x 10-5 ). HOCl(aq) (K a = 3.5 x 10-8 ). HCN(aq) (K a = 6.2 x ). HSO 4- (aq) (K a = ). H 2 O(l) (K W = 1.0 x ). 9 Aqueous Weak Bases (K b ~ 10 -x ): - remain UNDISSOCIATED as initial species. An base that is isn t strong is presumed weak (can always check the K b value ~ 10 -x ). Some Examples ( undissociated initially ): NH 3 (aq) (K b = 1.8 x 10-5 ). (C 2 H 5 )NH 2 (aq) (K b = 5.6 x 10-4 ) [ a PRIMARY (1º) amine - R-NH 2 ] (C 2 H 5 ) 2 NH(aq) (K b = 1.3 x 10-3 ) [ a SECONDARY (2º) amine - R 2 NH] (C 2 H 5 ) 3 N(aq)) (K b = 4.0 x 10-4 ) [ a TERTIARY (3º) amine - R 3 N ] H 2 O(l) (K W = 1.0 x ). 10 Aqueous IONIC COMPOUNDS - SALTS : - completely dissociate as initial species. [Na +, K +, & NO 3 - salts are water-soluble.] Some Examples: (Recall the common polyatomic ions - Chap. 4) NaNO 3 (aq) --(completely)--> Na + (aq) + NO 3- (aq). CaCl 2 (aq) --(completely)--> Ca +2 (aq) + 2 Cl - (aq). NH 4 NO 3 (aq) --(completely)--> NH 4+ (aq) + NO 3- (aq). Strong Acids (HX) versus Weak Acids (HA) C o of HX(aq): Initial species: H + (aq), X - (aq), H 2 O(l) [H + ] eq = C o. [Any H + (aq) from K W = (<< K a (HX)): H 2 O(l) <---> H + (aq) + OH - (aq) is subsequent and hence IGNORED by the MRA. (If C o 10-6 M at 25ºC. Recall that [H + ] eq = [OH - ] eq = 10-7 M in pure H 2 O at 25ºC. ) 11 12

4 Strong Acids (HX) versus Weak Acids (HA) C o of HA(aq) - WEAK acid (K a (HA) = 10 -x but still > K W ) Initial species: HA(aq) & H 2 O(l) Reactions: HA(aq) + H 2 O(l) <---> H 3 O + (aq) + A - (aq) ; K a > K W MAIN RXN is HYDROLYSIS of HA(aq) Strong Acids (HX) versus Weak Acids (HA) HA(aq) + H 2 O(l) <---> H 3 O + (aq) + A - (aq) ; K a [I] C o 0 0 [ ] -x +x +x [eq] C o - x +x +x K a = (x 2 ) / (C o - x) Do the math Solve exactly using quadratic formula [p. 2]. Neglect x w/r to C o - check with the 5 % rule MRA - Strong Acids (HX) vs Weak Acids (HA) HA(aq) + H 2 O(l) <---> H 3 O + (aq) + A - (aq) ; K a If neglect x : x = [H + ] eq = [A - ] eq = (K a C o ) 1/2 A useful quantity (p. 2) is % dissociation (of HA): % dissociation = (moles HA dissoc.) / (initial moles HA) = (100 %) [HA] dissoc / [HA] o = (100 %) x / C o (100 %) (K a C o ) 1/2 / C o (100 %) K a 1/2 C o -1/2 For a STRONG ACID (HX): % dissociation = 100 % MRA - Weak Acid (HA) [page 3] Example: C o of (aq) acetic acid (HOAc(aq)), K a = 1.8 x initial species: HOAc(aq) & H 2 O(l). HOAc(aq) + H 2 O(l) <---> H 3 O + (aq) + OAc - (aq) C o [H + ] eq % dissoc. ph eq

5 Strong (MOH or M(OH) 2 ) versus Weak(B:) Base [pages 3-5 of Acid-Base Handout] C o of MOH(aq) or M(OH) 2 (aq) Initial species: M + (aq) (or M +2 (aq)), OH - (aq), H 2 O(l) [OH - ] eq = C o (for MOH) or 2 C o (for M(OH) 2 ) [Any OH - (aq) from K W = (<< K b (s. base)): H 2 O(l) <---> H + (aq) + OH - (aq) is subsequent and hence IGNORED by the MRA. (If C o 10-6 M at 25ºC. Recall that [H + ] eq = [OH - ] eq = 10-7 M in pure H 2 O at 25ºC. ) Strong Bases versus Weak Bases [pages 3-5 of Acid-Base Handout] C o of B(aq) - WEAK base (K b (B) = 10 -x but still > K W ) Initial species: B(aq) & H 2 O(l) Reactions: B(aq) + H 2 O(l) <---> BH + (aq) + OH - (aq) ; K b > K W MAIN RXN is HYDROLYSIS of B(aq) Strong Bases versus Weak Bases [page 4 of Acid-Base Handout] B(aq) + H 2 O(l) <---> BH + (aq) + OH - (aq) ; K b [I] C o 0 0 [ ] -x +x +x [eq] C o - x +x +x K b = (x 2 ) / (C o - x) Do the math Solve exactly using quadratic formula [p. 4]. Neglect x w/r to C o - check with the 5 % rule. MRA - Strong Bases vs Weak Bases [page 4 of Acid-Base Handout] B(aq) + H 2 O(l) <---> BH + (aq) + OH - (aq) ; K b If neglect x : x = [OH - ] eq = [BH + ] eq = (K b C o ) 1/2 A useful quantity (p. 4) is % ionization (of B): % ionization = (moles B ioniz.) / (initial moles B) = (100 %) [B] ioniz / [B] o = (100 %) x / C o (100 %) (K b C o ) 1/2 / C o (100 %) K b 1/2 C o -1/2 For a STRONG BASE (MOH or M(OH) 2 ): % ionization = 100 % 19 20

6 MRA - Weak Base (B) [page 5] Example: C o of (aq) ammonia (NH 3 (aq)), K b = 1.8 x initial species: NH 3 (aq) & H 2 O(l). NH 3 (aq) + H 2 O(l) <---> NH 4+ (aq) + OH - (aq) C o [OH - ] eq % dissoc. ph eq MRA - Some More Examples [page 10 of Acid-Base Handout] A mixture of acids: 300 ml of 2.00 M HF(aq) (K a = 7.2 x 10-4, weak) mixed with 200 ml of 3.00 M HOCl(aq) (K a = 3.5 x 10-8, weak) Initial species: HF(aq), HOCl(aq), & H 2 O(l) Reactions: HF(aq) + H 2 O(l) <---> H 3 O + (aq) + F - (aq) ; K a (HF) HOCl(aq) + H 2 O(l) <---> H 3 O + (aq) + OCl - (aq) ; K a (HOCl) MAIN RXN is HYDROLYSIS of HF(aq) MRA - Some More Examples - [page 10] HF(aq) + H 2 O(l) <---> H 3 O + (aq) + F - (aq) ; K a [I] 1.20 M* 0 0 [ ] -x +x +x [eq] x +x +x Concentrations AFTER MIXING (vol s. additive): *[HF] o = (2 M) (300 ml)/(500 ml) = 1.20 M [HOCl] o = (3 M) (200 ml)/(500 ml) = 1.20 M K a = 7.20 x 10-4 = (x 2 ) / ( x) Do the math see pages MRA - Some More Examples - [pages 10-11] Equilibrium Concentrations [H + ] eq = [F - ] eq = x = M [HF] eq = x = M ph eq = -log 10 (0.0294) = 1.53 (acidic). WE WOULD BE DONE (as far as ph eq goes) but we are also asked to determine [HOCl] eq, [OCl - ] eq, & [OH - ] eq We continue The next most important reaction is the hydrolysis of HOCl(aq) [K a = 3.5 x 10-8 > K W ]

7 MRA - Some More Examples - [pages 11-12] We now set up the table with the any unreacted initial conditions & ANY EQUILIBRIUM CONCENTRATIONS FROM THE MAIN RXN. Secondary Reaction: HOCl(aq) + H 2 O(l) <---> H 3 O + (aq) + OCl - (aq) ; K a [I] 1.20 M* M 0 [ ] -x No - MRA +x [eq] x M +x Concentrations AFTER MIXING (vol s. additive): *[HOCl] o = (3 M) (200 ml)/(500 ml) = 1.20 M K a = 3.5 x 10-8 = (+x) (0.0294)/( x).. solve MRA - Some More Examples - [page 12] Equilibrium Concentrations Secondary Reaction: [OCl - ] eq = 1.43 x 10-6 M [HOCl] eq = x 1.20 M WE STILL HAVE ONE MORE SPECIES: [OH - ] eq. The last equilibrium is K W = 1.0 x ]. [I] Some M 0 [ ] No - MRA +x [eq] M +x K W = 1.0 x = (0.0294) (x) solve [OH - ] eq = 3.40 x M DONE!! 25 26