1. Weak acids. For a weak acid HA, there is less than 100% dissociation to ions. The B-L equilibrium is:

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1 th 9 Homework: Reding, M&F, ch. 15, pp , (clcultions of ph, etc., for wek cids, wek bses, polyprotic cids, nd slts; fctors ffecting cid strength). Problems: Nkon, ch. 18, #1-10, 16-18, 31, 33, 34, 39-42; M&F, ch. 15, #51, 57, 63, 73. V. Aqueous Equilibri: Acids nd Bses. E. ph nd other clcultions for wek cids nd bses. 1. Wek cids. For wek cid HA, there is less thn 100% dissocition to ions. The B-L equilibrium is: HA H2 O º H3O A A is the conjugte bse of HA. Ex. HF/F By convention, we omit the H2O on both sides, nd write: HA º H A K = K = cid-dissocition constnt = {[H ][A ]}/[HA] c Wek cids hve K vlues less thn or equl to 1. For strong cids, K >> 1. Ex. K (HF) = Ex. K (CH3 COOH) = Ex. K (HCN) = Since K is usully smll, only smll frction of HA dissocites to H nd A As with generl equilibri, there re mny types of problems you should be ble to solve. They will lso be orgnized by types. Type 1. The concentrtion nd ph of wek cid solution re given. Clculte the K of the wek cid. Assumption (vlid in this course): the dominnt source of H is the wek cid HA, not wter dissocition. Vritions of this ssumption will be used on ll cid-bse equilibri. Ex. The ph of 0.10 M HOCl solution is Clculte K (HOCl). Write the chemicl equilibri nd the equilibrium expression: HOCl º H OCl K = {[H ][OCl ]}/[HOCl] 0.10 x 0 x 0 x Use shortcut; set up tble (initil, ), & finl conc.) s one line. Let x = finl conc. of H. The initil conc. of H is 0 (see ssumption), nd )[H ] = x. By stoichiometry, )[OCl ] = x, [OCl ] = x, )[HOCl] = x, nd [HOCl] = 0.10 M x. In this cse, we cn get x from the ph x = [H ] = 10 = M. Hence, [OCl ] = M nd [HOCl] = M = 0.10 M 8 K = {[H ][OCl ]}/[HOCl] = {( )( )}/(0.10) = (2 sig. fig.) -31-

2 Introduce here percent dissocition = 100%@[A ]/C HA, where C HA is the forml concentrtion of HA (text clls it the initil concentrtion). The 0.10 M bove is the forml conc. of HOCl. Percent dissocition = 100%( )/(0.10) = 0.059% (not very much dissocition) Type 2. Given the K nd forml conc. of wek cid, clculte ph nd concentrtions of ll species in solution (importnt in environmentl chemistry cid rin). Ex. Clculte the ph nd conc. of ll species in 1.00 M cetic cid solution. K (CH3COOH) = Let HA = CH3 COOH (cetic cid) nd A = CH3COO (cette) HA º H A K = {[H ][A ]}/[HA] 1.00 x 0 x 0 x Following the sme procedure s for type 1, list initil, chnge nd finl conc. of ll species on one line. Recll ssumption (HA is only source of H ). Plug finl concentrtions into equilibrium expression nd solve for x; then solve for ll finl conc. nd ph = {(x)(x)}/(1.00 x) = x /(1.00 x) This is qudrtic eqution nd cn be solved using the qudrtic formul. However, lwys try this shortcut first. Since K is very smll, ssume x is smll compred to the forml conc.: = x /(1.00 x). x /(1.00) (x << 1.00) 2 ½ 3 x = (1.00)( ) = ; x = ( ) = Alwys check the ssumption. Use Finkle s 10% rule. If x < 10% of C HA, then you hve the finl solution for x. If x $ 10% of C HA, use the qudrtic formul. 3 3 Since < (0.1)(1.00) = 0.10, then x = is the finl solution. 3 3 [H ] = [A ] = M; ph = log( ) = [HA] = = 1.00 M; percent dissocition = 100%( )/(1.00) = 0.42% [OH ] = K w/[h ] = / = M The ptterns in this type of problem pper in mny similr problems to be discussed. Another exmple is shown to emphsize the steps. 4 Ex. Wht is the ph of 0.25 M HF? K (HF) = HF º H F K = {[H ][F ]}/[HF] 0.25 x 0 x 0 x -32-

3 = (x)(x)/(0.25 x). x /(0.25) (ssumption: x << 0.25) x = (0.25)( ) = ; x = (is < (0.1)(0.25)? yes) 3 3 [H ] = x = ; ph = log( ) = Wek polyprotic cids. Polyprotic cids contin more thn one dissocible proton. Common polyprotic cids include crbonic cid (H2CO 3 ), phosphoric cid (H3PO 4 ), nd sulfuric cid (H2SO 4). Polyprotic cids dissocite stepwise (1 H t time), ech step with its chrcteristic K : Ex. Crbonic cid - diprotic cid 7 H2CO 3 º H HCO3 K 1 = = {[H ][HCO 3 ]}/[H2CO 3] HCO 3 º H CO3 K 2 = = {[H ][CO 3 ]}/[HCO 3 ] K 1 6 loss of first H ; K 2 6 loss of second H ; nd so on. Becuse it is hrd to seprte opposite chrges (H from n nion), the following trend is generlly true: K >> K >> K Type 3. Given the K s nd the forml conc. of wek polyprotic cid, clculte ph nd concentrtions of ll species in solution. Becuse of the trend in K s noted bove, solution contining polyprotic cid cn be treted like monoprotic cid when clculting ph; use K. 1 2 Ex. Clculte the ph nd conc. of ll species in 0.50 M H2SO 3 solution. K 1 = nd 8 K 2 = Use just the first dissocition to get [H ] nd ph. H2SO 3 º H HSO3 K 1 = {[H ][HSO 3 ]}/[H2SO 3] 0.50 x 0 x 0 x = x /(0.50 x). x /0.50 (ssumption: x << 0.50) x = (0.50)( ) = ; x = (is x < (0.1)(0.50)? NO) 2 2 This one will require the full qudrtic solution. Solve: x = ( )(0.50 x) 2 Choose the positive root: x = M = [H ] = [HSO 3 ]; ph = 1.10 [H2SO 3] = = 0.42 M Wht bout the second dissocition? 2 HSO 3 º H SO3 K 2 = {[H ][SO 3 ]}/[HSO 3 ] x x 0 x 2 Assumption (lwys vlid in this course): x << (whtever [H ] is in bove clcultion) -33-

4 K 2 = {( )(x)}/( ) = x = M = [SO 3 ] Do you see the pttern now? Next, 3. Wek bses For wek bse B (where B = NH or ny other mine), the B-L equilibrium is: 3 B H O º BH OH K = bse-dissocition product = {[BH ][OH ]}/[B] 2 b Type 2': Given K nd the forml conc. of wek bse, clculte the ph. This problem is very b similr to type 2 (wek cid), with some simple substitutions. Ex. Wht is the ph of 0.40 M NH 3 solution? K b(nh 3) = NH 3 H2 O = NH 4 OH K b = {[NH 4 ][OH ]}/[NH 3] 0.40 x 0 x 0 x (let x = [OH ]) = (x)(x)/(0.40 x). x /(0.40) (ssumption: x << 0.40) x = (0.40)( ) = ; x = (is x < (0.1)(0.40)? yes) 3 3 x = [OH ] = [NH 4 ] = M; [NH 3] = = 0.40 M NOTE: there is n extr step needed to clculte ph 3 poh = log( ) = 2.57; ph = = Relity check: solution of bse should hve ph bove 7. Percent dissocition of bse = ]/C B, where C B is the forml conc. of bse. 3 In this cse, percent dissocition = 100% ( )/(0.40) = 0.68% 4. Reltionship between K nd K b for conjugte cid/bse pir. Recll erlier sttement: if the conjugte cid is wek, the conjugte bse is wek, nd vis vers. Consequence: HF is wek cid; hence, F is wek bse. Note this rule: K mens tht H is product, K b mens tht OH is product. Ex. HF º H F K = {[H ][F ]}/[HF] F H2 O º HF OH K b = {[HF][OH ]}/[F ] (clled the hydrolysis rxn) The conjugte bse F rects with wter to produce the conjugte cid nd OH. Solutions of NF re bsic (ph > 7). (comment: grsping this concept ppers to be one of the hrdest prts of cid/bse theory for mny students. It is relly quite simple) Note: Nkon clls K the b hydroysis constnt. -34-

5 Ex. NH 3 H2 O º NH 4 OH K b = {[NH 4 ][OH ]}/[NH 3] NH 4 º NH 3 H K = {[NH 3][H ]}/[NH 4 ] The conjugte cid NH 4 dissocites to produce the conjugte bse nd H. Solutions of NH4Cl re cidic (ph < 7). You cn now do ph clcultions of cidic or bsic slts; they follow exctly the sme ptterns s wek cids (type 2) nd wek bses (type 2'). Exmples pper in the homework; plese work them. You need one other fct. For ny conjugte cid/bse pir: K@K = K b w Proof given in the text. This is mthemticl expression of the trend: the stronger the conjugte cid (lrger K ), the weker the conjugte bse (smller K ), nd vis vers. b Ex. K (HF) = ; wht is K b for F? K b = K w/k = / = F. Fctors tht ffect cid strength. This mteril will not be covered in lecture. Plese remember the following trends. 1. Binry cids H A nd H B, where A & B re 2 elements on column in the Periodic n n Tble: cid strength increses down the column (weker H-B bond). Ex. HF is weker thn HCl. 2. Binry cids H A nd H B, where A & B re 2 elements cross row in the Periodic n n Tble: cid strength increses from left to right (incresing electronegtivity of B). Ex. H O is weker thn HF Oxocids H XO nd H YO, where n nd m re the sme: the more electronegtive n m n m centrl tom yields the stronger cid. Ex. HOBr is weker thn HOCl. 4. Oxocids H XO nd H YO, where n is the sme nd o > m: the more oxygens, the n m n o stronger the cid. Ex. HOCl is weker thn HClO