122 HW_17 key

122 HW_17 key - 17.2 17.5 \ 17.7 \: /Z..W/ £// 7 If an...

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Unformatted text preview: 17.2 17.5 \ 17.7 \: /Z..W/ £// 7 If an external source of BH‘ such as BH*C|‘ is added to a solution of B(aq), [BH*] increases, decreasing [OH'] and \‘ (a) For a generic weak base B, Kb = increasing [B], effectively suppressing the ionization (hydrolysis) of B. (b) NH4Cl Analyze/Plan. Follow the logic in Sample Exercise 17.1. HCaH502(aQ) : H*(aq) + 03H502‘(aq) 0.060 M +x (0.060 + x) M i 0.085 M c -x +x e (0.085 — x) M +X M [H ‘1 [03H502J1 = (x)(0.060+ x) =1. x10‘5= K“ 3 [HC3H5OZ] (0.085—x) Assume x is small compared to 0.060 and 0.085. 1.3 x104: O‘OGOX- x=1.8 x 10*: [H*]. pH =4.73 0.085 [03H502‘] = 0.060 + 1.8 X 10‘5 = 6.002 X 10'2 = 6.0 x 10‘2 M Analyze/Plan. Follow the logic in Sample Exercise 17.1. This exercise is more straightforward than the sample, because we are given concentrations directly. Solve: (a) HCHOZ is a weak acid, and NaCHO2 contains the common ion CHOz', the conjugate base of HCHOZ. Solve the common-ion equilibrium problem. HCHOZ(aq) # H*(aq) + CH0;(aq) i 0.250 M 0.160 M c -X +x +x e (0.260 - X) M +X M (0.160 + X) M -18 1 0.160x K3 ' " ° ‘ [HCH021 (0.260 -x> 0.260 x = 2.93 x10"4 = 2.9 X10‘4 M: [H‘], pH = 3.53 Check Since the extent of ionization of a weak acid or base is suppressed by the 4/\ \& presence of a conjugate salt, the 5% rule usually holds true in buffer ‘\ {4 solutions. (b) CsHsN is a weak base, and C5H5NHCl contains the common ion C5H5NH*, which is the conjugate acid of C5H5N. Solve the common ion equilibrium problem. CsHsN(aQ)+H20(l) ‘1‘ CsHsNH+(aQ) + 0H’(aQ) i 0-210M 0.350M c 'x +X +x 6 (0-210-x)M (0.0350+x)M +xM -1 7 x 10.9- [CSHSNH +1 [0” "l _ N 0.3sz K” ' [05Hle (0.210-x) ” 0210 x = 1.02 X10'9 = 1.0 X10"9 M= [OH‘], pOH = 8.991, pH = 14.00 - 8.991 = 5.01 Check. In a buffer, if [conj. acid] > [conj. base], pH < pKa of the con]. acid. If [conj. acid] < [conj. base], pH > pKa of the con]. acid. in this buffer, pKa of CSHSNH+ is 5.23. [C5H5NH*] > [CsHsN] and pH = 5.009, less than 5.23. "17.13 Analyze/Plan. Follow the loci: Sample Exercise 17.3. Assume that % ionization is small in these buffers (Solutions 17.9 and 17.10). Solve: K3: [H *][Lac ‘1 ; [Ht]: [Kai [ii—ac] = 1.4x 10‘4 (0.12) [HLac] [Lac ] (0.11) .. [H*]=1.53X10'4=1.5X10‘4;pH=3.82 (a) (b) mol = M x L; total volume = 85 mL + 95 mL = 180 mL K [HLaC] = 1.4 x 10"4 (0.13 Mx 0.085 L)/O.18O L = 1.4 x 10‘4 (0.13 x 0.085) [Lac 1 (0.15 Mx 0.095 L)/0.18O L (0.15 x 0.095) [H+]= 1.086 X10"“=1.1MH*;pH =3.96 . [H*][CHO’] K [HCHQJTT I (a) Ka = 3 5 2 ; = a 3 .5~ 2 ,_ I I [HCSHSOZ] [03H502] Since this expression contains a ratio of concentrations, we can ignore total volume and work directly with moles. 1.3 x 10‘5 (0.12 — x) z 1.3 x 10‘5 (0.12) [w] = (0.10 + x) ' 0,10 =1.56’<10‘5 =1.6><10‘5 M, pH=4.81 (b) HCaH502(aCI) + OH—(aCI) ")- CzH302_(aQ) + H200) 0.12 mol 0.01 mol 0.10 mol —0.01 moi -0.01 moi +0.01 mol 0.11 mol 0 mol 0.11 mol 1.3x10’5(0.11)=1 [H*] z .3 x 10‘5 M; pH = 4.89 (0.11) (c) 03H502_(aQ) + H1(aci) -> HCsH502(aQ) + I‘(aQ) 0.10 mol 0.01 mol 0.12 mol -0.01 mol -0.01 mol +0.01 mol 0.09 mol 0 mol 0.13 mol 1.3x 10'5(o.13) [H‘]: a1.88><10'3=2><10'3 M; pH=4.73=4.7 (0.09) 17.24;. (a) The quantity of base required to reach the equivalence point is the same in the two ’ ' titrations. (b) The pH is higher initially in the titration of a weak acid. (c) The pH is higher at the equivalence point in the titration of a weak acid. (d) The pH in excess base is essentially the same for the two cases. (e) In titrating a weak acid, one needs an indicator that changes at a higher pH than for the strong acid titration. The choice is more critical because the change in pH close to the equivalence point is smaller for the weak acid titration. ‘1 (a) HX is weaker. The pH at the equivalence point is determined by the identity and concentration of the conjugate base, X‘ or Y‘. The higher the pH at the equivalence point, the stronger the conjugate base (X‘) and the weaker the conjugate acid (HX). (b) Phenolphthalein, which changes color in the pH 8—10 range, is perfect for HX and probably appropriate for HY. Bromthymol blue changes from 6-7.5, and thymol blue between from 8—9.5, but these are two-color indicators. One-color indicators such as phenolphthalein are preferred because detection of the color change is more reproducible. 17.32 fa) (b-f) (b) (0) (d) (e) (f) (b) Weak base problem: K” = 1.8 x 10'5 = [NH4’] [OH '] [NH3] At equilibrium, [OH'] = x, [NH3] = (0.030 - x); [NHf] = x x2 x2 1.8 x 10'5 = —— z (0.030—X) 0.030 ; x= [OH‘] = 7.348 x 104 = 7.3 x 10“4 M pH= 14.00-3.13= 10.87 Calculate mol NH3 and mol NH4+ after the acid-base reaction takes place. 0.030 M NH3 x 0.0300 L = 9.0 x 10‘4 mol NH3 present initially. NH3(aq) + HCKaQ) -> NH4*(aq) + CHaCI) (0.025 M X 0.0100 L) = before rx 9.0 x 10“ mol 2.5 x 10‘4 mol . 0 mol after rx 6.5 x 10“ mol 0 mol 2.5 X 10“ mol (0.025 M x 0.0200 L) = before rx 9.0 x 10“ mol 5.0 x 10‘4 mol 0 mol after rx 4.0 x 10“ mol 0 mol 5.0 x 10" mol (0.025 M x 0.0350 L) = before rx 9.0 x 10“ mol 8.75 x 10‘4 mol 0 mol after rx 0.25 x 10“ mol 0 mol 8.75 x 10" mol (0.025 M X 0.0360 L) = before rx 9.0 x 10" mol 9.0 x 10" mol 0 mol after rx 0 mol 0 mol 9.0 x 10" mol (0.025 M X 0.0370 L) = before rx 9.0 x 10‘4 mol 9.25 x 10‘4 mol 0 mol after rx 0 mol 0.25 x 10‘4 mol 9.0 x 10" mol - Using the acid dissociation equilibrium for NH4+ (so that we calculate [H*] directly), NH +(aQ) # HWEQ) + NH3(aQ) 4 H’ NH K ‘14 K3: [ H 3] = W = 10"“) =5,56><10'1° = 5.6x10‘1° —4 4 [NH3] = has X 10 m0' = 0.01625 M; [NHf] = —____2'50 x10 “‘0' = 6.25 x 104 M 0.0400 L 0.0400 L 5.56 x 10'10 [NH ‘1 -10 .3 H = ——~—“ 6' M556 x ‘0 (525x10 )=2.14 x 104°; pH = 9.67 [NH3] (0.01625) (We will assume [H*] is small compared to [NH3] and [NH4*].) _4 ‘4 m = (10030 M; [NH4+] = w = 0,010M NH = (c) I 3] 0.0500L 0.0500L 710 [H‘] = w = 6.94 x 10-10 = 6.9 x10-1° M; pH = 9.16 (0.0080) 24 —4 (d) [NH3] = 0.25 x 10 mol = 3846 x104 = 4 x1o_4M;[NH4+]= 8.75x 10 mol 0.0650 L 0.0650 L = 0.01346 = 0.013 M '10 [H] = =1,946 x1()“'8 :2 x 10-5 M; pH :7] 3.846 x 10’4 (e) At the equivalence point, [H“] = [NHe] = x —4 [NH;] = W = 0.01364 = 0.014 M 0.0660 L 2 5.56 X104" = X ; x = [H1 =2.754 X10"6 =2.8 X1045; pH = 5.56 0.01364 (f) Past the equivalence point, [H‘] from the excess HCl determines the pH. + 0.25 x 10'4mol [H ]: _____ = 3.731 X10‘4 = 4 x10“4 M; pH = 3.4 0.0670L 17.33 Analyze/Plan. Calculate the pH at the equivalence point for the titration of several bases with 0 ' 0.200 M HBr. The volume of 0.200 M HBr required in all cases equals the volume of base and the final volume = 2Vbase. The concentration of the salt produced at the equivalence point is 0.200 M x V vaase In each case, identify the salt present at the equivalence point, determine its acid-base properties (Section 16.9), and solve the pH problem. Solve: W = 0.100 M. (a) NaOH is a strong base; the salt present at the equivalence point, NaBr, does not affect the pH of the solution. 0.100 M NaBr, pH = 7.00 L (b) HONH2 is a weak base, so the salt present at the equivalence point is HONHJBr'. Ill) » This is the salt of a strong acid and a weak base, so it produces an acidic solution. l ‘v ‘ C 0.100 M HONH3*Br‘; HONH3"(aq) r—‘ H“(aq) + HONH2 [equil] 0.100 - x x x H+ HONH K -14 Ka=——-—[ M 2] =—w=————-1'0X1° =9.09x10-7=9.1x10-7 [HONHg] Kb 1 1 x 10'8 Assume x is small with respect to [salt]. Ka = x2/0.100; x = [H1 = 3.02 ><10‘4 = 3.0 X10“4 M, pH = 3.52 (c) CSHSNH2 is a weak base and CsHsNHa’Br' is an acidic salt. 0.100 M C6H5NH3“Br' . Proceeding as in (b): _ [H *] [CGHSNHZ] _ K [ceHSNHQ Kb [H*]2 = 0.100(2.33 x 104 ); [H‘] = 1.52 x 10'3 = 1.5 3‘10"3 M, pH = 2.82 ...
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This note was uploaded on 07/17/2008 for the course CHEM 122 taught by Professor Zellmer during the Winter '07 term at Ohio State.

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122 HW_17 key - 17.2 17.5 \ 17.7 \: /Z..W/ £// 7 If an...

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