C deposits (63.5/2) g of copper; the quantity passed is therefore

Transcription

1 7. SOLUTIONS OF ELECTROLYTES n Faraday s Laws, Molar Conductivity, and Weak Electrolytes C deposits (63.5/2) g of copper; the quantity passed is therefore C The current was passed for 3600 s; the current is therefore A = A = 33.8 ma 7.2. Quantity of electricity passed = C Current = A = A = 2.38 ma 2.4 ma 7.3. The chemical reactions involved are C 6 H 5 OH + Br 2 (g) C 6 H 4 (Br)OH + HBr 2Br Br 2 (g) + 2e 2K + + 2e 2K(s) Two moles of electrons are involved in the generation of each mole of bromine gas, which reacts with one mole of phenol. Each batch consists of kg ( g) or 5313 moles of phenol (molecular weight = g mol 1 ), which requires 5313 mol. of bromine, or mol. of electrons for the reaction. Therefore, since It = nf, where n is the number of moles of electrons exchanged in the reaction, we get mol C mol 1 1 h t = C s s = 14.3 h.

2 SOLUTIONS OF ELECTROLYTES n c/10 4 M Λ α 1 α K = cα 2 /(1 α) Ω 1 cm 2 mol M The values are reasonably constant; average K = mol dm Λ AgCl = = Ω 1 cm 2 mol 1 Solubility = mol cm 3 = mol cm 3 = mol dm 3 = 9.11 µm 7.6. The increase in conductivity, Ω 1 cm 1, is due to the CaSO 4 present; thus Λ 1 2 Ca SO2 4 = Ω 1 cm 1 2c where c is the concentration of CaSO 4 ; 2c is the concentration of 1 2 CaSO 4. The value of λ 1 2 SO2 4 is obtained from the conductivity of the Na 2 SO 4 solution: Λ Na SO2 4 = Ω 1 cm mol cm 3 (Note that since the concentration of Na 2 SO 4 is M, that of 1 2 Na 2 SO 4 is M.) Thus Λ Thus, since Then Na SO2 4 λ(na + ) = 50.1 Ω 1 cm 2 mol 1 λ Λ 1 2 SO Ca SO2 4 = Ω 1 cm 2 mol 1 = 79.9 Ω 1 cm 2 mol 1 = Ω 1 cm 2 mol 1

3 138 n CHAPTER 7 and c = Thus Ω 1 cm Ω 1 cm 2 mol 1 = mol dm 3 c Ca 2+ = mol dm 3 c SO 2 4 = ( ) mol dm 3 = mol dm 3 K sp = mol 2 dm From Eq. (7.9) we have Λ = κ/c; therefore the electrolytic conductance κ is κ = Λc = S cm 2 mol mol cm 3 = S cm 1. Now, using Eq. (7.8), the cell constant is l/a = κ /G = S cm S = cm 1. To find the equivalent conductance of the electrolyte, we use Λ = G(l/A) c = cm S S = 180 S cm 2 mol Λ(KCl) = ( ) Ω 1 cm 2 mol 1 The electrolytic conductivity at 0.01 M is κ(kcl) = Ω 1 cm 2 mol mol cm 3 = Ω 1 cm 1 Recall that conductance is inversely proportional to the resistance. The electrolytic conductivity of the ammonia solution is thus κ(nh 4 OH) = = Ω 1 cm 1 The molar conductivity of NH OH is Λ(NH OH ) = ( ) Ω 1 cm 2 mol 1 If c = [NH + 4 ] = [OH ], Ω 1 cm 2 mol 1 = Ω 1 cm 1 c c = mol cm 3 = mol dm 3

4 SOLUTIONS OF ELECTROLYTES n 139 The concentrations of NH 4 OH, NH + 4, and OH are thus NH 4 OH NH OH mol dm 3 K b = mol dm From the conductivity and concentration, we get Λ = 4 S cm mol cm 3 = 4.90 S cm2 mol 1. For the weak base, we write B + H 2 O BH + + OH c(1 α) αc αc so that K b = [BH+ ][OH ] [B] = α 2 c (1 α). Since α = Λ/Λ = , from which with c = mol dm 3, we calculate K b = ( ) mol dm 3 ( ) = mol dm Note that each number in the first row (concentrations) must be multiplied by 10 3 M to yield the molar concentration. Using the model suggested by the Debye-Huckel Onsager equation (Equation 7.53), we assign equivalent conductance as the dependent variable and c as the independent variable. The result of the linear regression is Λ = c. In the limit as c 0, we have Λ = S cm 2 mol Equation 7.20 can be rearranged to cλ 2 = KΛ 2 0 KΛ 0 Λ cλ 2 could therefore be plotted against Λ. Alternatively, since cλ = KΛ Λ KΛ 0 cλ can be plotted against 1/Λ. The slope and intercepts are as shown below:

5 140 n CHAPTER 7 cλ 0 KΛ 0 1/Λ 0 2 Slope = KΛ 0 1/Λ Λ values are obtained by the use of Eq. 7.9; for the lowest concentration, mol dm 3, Λ = Ω 1 cm dm 3 cm mol dm 3 = 11.4 Ω 1 cm 2 mol 1 Similarly, for the other concentrations: c/10 4 mol dm Λ/Ω 1 cm 2 mol cλ/10 6 Ω 1 cm 1 mol /(Λ/Ω 1 cm 2 mol 1 ) In a plot of cλ against 1/Λ, the intercepts are KΛ 0 = Ω 1 cm 1 mol 1 1/Λ 0 = Ω 1 cm 2 mol 1 ; Λ 0 = 30 Ω 1 cm 2 mol 1 K = mol cm 3 = mol dm The concentration of the acid in water = 1500 ppm = g acid = 10 3 g solution = m g acid 10 6 g solution = g acid/60.05 g mol kg solution Since the solution has the same density as water, 1.00 kg of solution has a volume of 1.0 dm 3. In other words, we may assume the solution to have a concentration of M. Now, for a weak acid whose degree of dissociation is α and the concentration is c M, Eq. (7.18) gives α K a = 2 c (1 α), or α2 c + K a α K a = 0. Solving this equation for the degree of dissociation α with c = M gives α = (the other solution is negative). Since α = Λ/Λ, we have Λ = αλ = S cm 2 mol 1 = S cm 2 mol 1.

6 SOLUTIONS OF ELECTROLYTES n 141 Therefore, the conductance measured by the cell cannot exceed Λc = S cm 2 mol mol cm 3 = S cm The Λ value for H 2 O is calculated as Λ (HCl) Λ (KCl) + Λ (KOH) = S cm 2 mol 1. In pure water, the only species conducting electricity are H + and OH ions, each of which have concentrations of = mol dm 3. Since this is a very low concentration, we may assume that Λ Λ. Therefore, κ = Λc = mol cm S cm 2 mol 1 = S cm 1. n Debye-Hückel Theory and Transport of Electrolytes From Eq. 7.50, Thickness c 1/2 Thickness ε 1/2 Therefore, a. At M, thickness = = 30.5 nm b. At ε = 38, thickness = = nm Λ 1/2Na2 SO 4 = Λ NaCl + Λ 1/2K2 SO 4 Λ KCl = = Ω 1 cm 2 mol Λ NH4 OH = Λ NH 4 Cl λ Cl + λ OH α = = = cm 2 Ω 1 mol =

7 142 n CHAPTER a. Quantity of electricity = 2 h 3600 s h A = 5688 C Amount deposited = 5688/ = mol g Loss of LiCl in anode compartment = g mol 1 = mol Anode reaction: Cl 1 2 Cl 2 + e mol Cl is removed by electrolysis. Net loss = mol Cl = = mol Cl have migrated into the anode compartment. t Cl = = t Li + = = b. λ Li + = = 36.5 Ω 1 cm 2 mol 1 λ Cl = 78.5 Ω 1 cm 2 mol 1 Then, from Eq. 7.64, u + = 36.5/ = cm 2 V 1 s 1 u = 78.5/ = cm 2 V 1 s Molecular weight of CdI 2 = C deposits 1 2 mol Cd2+ = g of Cd current passed is = C Anode compartment ( g) originally contained It finally contains = mol = mol

8 SOLUTIONS OF ELECTROLYTES n 143 Loss in anode compartment = mol C would have brought about a loss of = mol of CuI 2 = mol of 1 2 CuI 2 t + = ; t = The individual ionic conductivities are λ + = = Ω 1 cm 2 mol 1 λ = = 76.3 Ω 1 cm 2 mol 1 Then, by Eq. 7.64, the ionic mobilities are u + = Ω 1 cm 2 mol C = cm 2 V 1 s 1 u = 76.3 Ω 1 cm 2 mol C = cm 2 V 1 s The ionic mobilities are (Eq. 7.64) u + = 50.1 Ω 1 cm 2 mol C = cm 2 V 1 s 1 u = 76.4 Ω 1 cm 2 mol C mol 1 = cm 2 V 1 s 1 The velocities in a gradient of 100 V cm 1 are thus Na + : cm s 1 Cl : cm s The molar conductivity of LiCl is Λ = ( ) Ω 1 cm 2 mol 1 The specific conductivity of a 0.01 M solution is this quantity multiplied by 10 4 mol cm 3 : κ = Ω 1 cm 1 The resistance of a l-cm length of tube is thus 1 cm/5 cm R = Ω 1 cm 1 = Ω The potential required to produce a current of 1 A is Ω 1 A = V The potential gradient is thus V cm 1.

9 144 n CHAPTER 7 The mobilities of the ions are (Eq. 7.64) Li + : Cl : 38.6 Ω 1 cm 2 mol C mol 1 = cm 2 V 1 s Ω 1 cm 2 mol C mol 1 = cm 2 V 1 s 1 The velocities are Li + : cm s 1 ; Cl : cm s The work is given by dw = Fdr, where the force of attraction is F = Q 1 Q 2 /r 2 Therefore w = r 1 Q 1 Q 2 4πε 0 r 2 dr = Q 1 Q 2 4πε r 1 a. ε 0 = C 2 J 1 m 1 ; r 1 = 10 9 m w = ( C) 2 4π C 2 J 1 m m = J b. w = (1/ 1/10 3 m) = J 1 c. w = (1/0.10 m 1/10 9 m) = ( ) = J The exponential is shown as curve a, 4πr 2 as curve b, and their product as curve c in the accompanying diagram. With z c = 1 and z i = 1, the function to be differentiated is f = e e2 /4πε 0 εrk B T 4pr 2 Differentiation gives df dr = 8πr /4πε ee2 0 εrk B T 4πr 2 e 2 4πε 0 εr 2 k B T ee2 /4πε 0 εrk B T Setting this equal to zero leads to r * = e 2 8πε 0 εk B T

10 SOLUTIONS OF ELECTROLYTES n 145 (a) (b) (c) 40 (a) e e2 /4πε O εrk B T (b) 4πr (c) e e2 /4πε O εrk B T 4πr a b 6 25 c r/nm r/nm The value of this at 25.0 C, with ε = 78.3, is m = nm With z c = 1, the potential energy for two univalent ions, from Eq. 7.47, is E p = e 2 4πε 0 εr Introduction of the expression for r * gives E p = 2k B T At 25.0 C, E p = J = 4.96 kj mol 1 n Thermodynamics of Ions NaCl: = kj mol 1 CaCl 2 : = kj mol 1 ZnBr 2 : ( ) = kj mol 1

11 146 n CHAPTER H + : kj mol 1 Na + : = kj mol 1 Mg 2+ : ( ) = kj mol 1 Al 3+ : ( ) = kj mol 1 Cl : = kj mol 1 Br : = kj mol KNO 3 : I = 1 2 ( ) = 0.1 M K 2 SO 4 : ZnSO 4 : ZnCl 2 : I = 1 2 ( ) = 0.3 M I = 1 2 ( ) = 0.4 M I = 1 2 ( ) = 0.3 M K 4 Fe(CN) 6 : I = 1 2 ( ) = 1.0 M Ionic strength of solution. I = 1 2 ( ) = 0.6 M log 10 γ ± = z + z 0.51 I = = = 1.58 γ ± = a. s = M log 10 γ ± = 0.51 ( ) 1/2 = γ ± = K s = γ 2 ± s2 = ( ) 2 = G = RT ln K s = ln = kj mol 1

12 SOLUTIONS OF ELECTROLYTES n 147 b. I = 1 2 ( ) = M log 10 γ ± = = γ ± = s = K s 1/2 = ( ) 1/2 γ ± = M z + z 0.51 I log 10 γ ± = 1 + a ( ) I For a = 0 and z + and z = 1 log 10 γ ± = 0.51 I I log 10 γ ± For a = 0.1 log 10 γ ± = 0.51 I I I log 10 γ ± For a = 0.2 log 10 γ ± = 0.51 I I I log 10 γ ± For a = 0.4 log 10 γ ± = For a = 0.8 log 10 γ ± = 0.51 I I 0.51 I I I log 10 γ ± I log 10 γ ±

13 148 n CHAPTER I log 10 γ ± a = 0.8 a = 0.4 a = 0.2 a = 0.1 a = 0 nm nm nm nm nm The electrostatic contribution to the Gibbs energy (Eq. 7.87) is, per mole of ions, G es = z2 e 2 L 8πε 0 εr = ( ) π ε J mol 1 = ε In the membrane, J mol 1 G es = kj mol 1 In water, G es = 6.7 kj mol 1 G es (water membrane) = 124 kj mol Λ 1 2 CaF 2 = = 98.1 Ω 1 cm 2 mol 1 Observed κ due to salt = = Ω 1 cm 1

14 SOLUTIONS OF ELECTROLYTES n 149 Solubility = mol cm 3 = mol dm 3 of 1 2 CaF 2 1 mol of 1 2 CaF 2 has a mass of = g. Solubility = g dm 3 Solubility product = [Ca 2+ ][F ] 2 = ( ) ( ) 2 = mol 3 dm x M CuSO 4 : I = 1 2 ( ) x = 4x M I = 0.1 M if x = x M Ni(NO 3 ) 2 : I = 1 2 (22 + 2) x = 3x I = 0.1 M if x = x M Al 2 (SO 4 ) 3 : I = 1 2 ( ) x = 15x M I = 0.1 M if x = x M Na 3 PO 4 : I = 1 2 ( ) x = 6x M I = 0.1 M if x = a. First, neglect the effect of activity coefficients: if s is the solubility s(2s) 2 = mol 3 dm 9 s = mol dm 3 The ionic strength is 1 2 ( ) = mol dm 3 By the Debye-Hückel limiting law log 10 γ ± = = γ ± = 0.88 If now the true solubility is s, the activities of the ions are Pb 2+ : γ + s; F : 2 γ s

15 150 n CHAPTER 7 Then (γ + s) (2γ s) 2 = mol 3 dm 3 γ + γ 2 4s3 = mol 3 dm 9 γ 3 ± 4s3 = mol 3 dm 9 (from Eq ) Thus s 3 = mol 3 dm 9 (0.88) 3 4 s = mol dm 3 We could proceed to further approximations as necessary. b. In 0.01 M NaF, the ionic strength is essentially 0.01 mol dm 3 and log 10 γ ± = = γ ± = If s is the solubility, s = [Pb 2+ ]; [F ] = 0.01 mol dm 3 Then sγ + (0.01 γ ) 2 = mol dm 3 γ + γ 2 s = mol dm 3 γ 3 ± s = mol dm 3 s = mol dm (0.791) 3 = mol dm We proceed by successive approximations, first taking the activity coefficients to be unity. Then, if s is the solubility, s 2 = mol 2 dm 6 s = mol dm 3 This is the ionic strength, thus log 10 γ ± = = γ ± = To a second approximation, γ 2 ± s2 = (0.744) 2 s 2 = mol 2 dm 6 s = mol dm 3

16 SOLUTIONS OF ELECTROLYTES n 151 To a third approximation, log 10 γ ± = ; γ ± = 0.71 (0.71) 2 s 2 = mol 2 dm 6 s = mol dm 3 To a fourth approximation, log 10 γ ± = ; γ ± = (0.704) 2 s 2 = s = mol dm For Problem 7.24 it was found that G es = ε J mol 1 For the transfer from water (ε 1 ) to lipid (ε 2 ) G es /J mol 1 = ε 2 ε 1 S es = G es T P Since ε 2 is temperature independent, this leads to (from Eq ) S es /J K 1 mol 1 = ε 1 T = ε 2 1 ε T = ε 1 ln ε T = = 31 J K 1 mol 1 The entropy increases because of the release of bound water molecules when the K + ions pass into the lipid.

17 152 n CHAPTER a. At infinite dilution the work of charging an ion is given directly by (Eq. 7.86): w rev = z 2 e 2 8πε 0 εr For 1 mol of Na + w rev = ( C) mol 1 8π C 2 N 1 m m = 9373 J mol 1 For 1 mol of Cl, w rev = 4920 J mol 1 For 1 mol of Na + Cl at infinite dilution, w rev = J mol 1 = 14.3 kj mol 1 b. These values are reduced when the electrolyte is at a higher concentration, the work of charging the ionic atmosphere being negative and equal to kt ln γ i. Thus, for 1 mol of Na + ions, of activity coefficient γ +, the work of charging the atmosphere is RT ln γ + Similarly for the chloride ion, the work per mole is RT ln γ For 1 mol of Na + Cl w rev (atm) = RT(ln γ + + ln γ ) = RT ln γ + γ = 2RT ln γ ± If γ ± = 0.70 w rev (atm) = 2( J mol 1 ) ln 0.70 = 1768 J mol 1 The net work of charging is thus w rev = = J mol 1 = 12.5 kj mol 1

18 SOLUTIONS OF ELECTROLYTES n The ionic strength of the solution is I = 1 2 [ (0.2 4) + 0.4] = 0.70 M The mean activity coefficient γ i of the barium and sulfate ions is given by log 10 γ i = = γ i = If the solubility in the solution is s, = s 2 (0.0196) 2 whence s = M The ionic strength of the solution is I = 1 2 [ ( )] = 0.03 M By the DHLL, log 10 γ i = = γ i = The solubility product is therefore K s = ( ) 2 (0.816) 2 = M 2 The solubility in pure water is thus ( ) 1/2 = M The enthalpy change H neut for the neutralization of HCN by NaOH is less than the value kj mol 1 because of the energy required for the dissociation of HCN, H diss, H neut = kj mol 1 H diss Thus H diss = kj mol 1 H neut = = kj mol 1

19 154 n CHAPTER I = 1 2 ( ) = From Eq , for Na + log γ i = z i 2 B I log γ Na + = = γ Na + = for SO 2 4 log γ 2 SO = = γ 2 SO = From Eq , γ ± = 0.51 z + z I = = = n Ionic Equilibria Palmitate side Other side Initial concentrations: [Na + ] = 0.1 M [Na + ] = 0.2 M [P ] = 0.1 M [Cl ] = 0.2 M Final concentrations: [Na + ] = (0.1 + x)m [Na + ] = (0.2 x)m [P ] = 0.1 M [Cl ] = (0.2 x)m [Cl ] = x M

20 SOLUTIONS OF ELECTROLYTES n 155 Then (0.2 x) 2 = (0.1 + x) x + x 2 = x x x = = 0.08 Final concentrations are thus, on the palmitate side, [Na + ] = 0.18 M; On the other side, [Cl ] = 0.08 M [Na + ] = [Cl ] = 0.12 M [H 2 NCH 2 COOH] [H + ] [H 3 N + CH 2 COOH] [H 3 N + CH 2 COO ] [H + ] [H 3 N + CH 2 COOH] = M = Dividing the first by the second gives [H 2 NCH 2 COOH] [H 3 N + CH 2 COO = ] = This is convincing evidence for the predominance of the zwitterion H 3 N + CH 2 COO a. ph = 1; H 3 PO 4 predominant b. ph = 2.7; H 2 PO 4 predominant c. ph = 4.3; H 2 PO 4 predominant d. ph = 11.4; HPO 2 4 predominant e. ph = 14; PO 3 4 predominant Let the final concentrations be Left-hand Compartment Right-hand Compartment [K + ]/M = 0.05 x [K + ]/M = x [Cl ]/M = 0.05 x At equilibrium, [Cl ]/M = x [P ]/M = 0.1 (0.05 x) 2 = x ( x) whence x = 0.01

21 156 n CHAPTER 7 The final concentrations are therefore Left-hand Compartment Right-hand Compartment [K + ] = 0.04 M [K + ] = 0.16 M [Cl ] = 0.04 M [Cl ] = 0.01 M [P ] = 0.1 M It is easy to check that the product [K + ][Cl ] is the same on each side of the membrane.