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Homework 10a

# Homework 10a - 324 Section 2 Solutions and Comments on...

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Unformatted text preview: 324 Section 2 Solutions and Comments on Problems 0.00001 0.917 0.00010 0.761 0.00100 0.422 9.11 This yields IOIDEU-‘lmethanol = 103-7343103Walwater = [IOIGBHIlwaterP‘mﬂ' so that (8} (yf)methanol = [(7,1)waterl3'7343 r (9) which is the required equation. Notice that the solution will be more nonideal in methanol than in H20 since Y: < 1 and we have (Vilmemami = [(74) 3.7343 < (7i)water- I water] An investigator determines the activity coefficients for a salt and finds the data in the table to the left. If the salt is represented by the molecular for- mula CHAL, what are the possible values of 12+ and 12,? Show that your answer is correct. Solution The Debye—I-Iiickel limiting law requires that, for low concentrations we have loghq) = —0.50926z+z_51*’2 = 70.509262+z_[0.5 m{vrzi + 22,21 ”1/2. {1) However, charge neutrality requires that we have 2 , = v_ and z _. = Lu. Inserting these equalities into Eq. 1 gives log(y,) = —0.36010v,V+[V+V2, + Lair-”2m”? (2) Rearranging terms in Eq. 2 produces . . logo.) 31212 _+p. 1“2=—. 3 V V [D ] —0.36010 m“2 ( ) Substituting the data yields vyzvi’q[v_ + 11+]U2 = ﬂ = 33.05, {4) —0.36010(0.00{J01)1~2 , 10 0.761 93:01)?“ + 15]”: = mggﬁ), = 32.94, (5) —0.36010{0.0001)“- and 10 0.422 ”Java/ﬁn. + vlrﬂ ’ *ﬂ—l : 32.90. (6) _ ~0.36010(0.001 )“2 The left—hand side of Eq. 3 is symmetric in 12+ and v_; therefore, the possibilities are as follows: for which we have va’iﬁleh + 14]”2 = 1.4142 . . . ,' 14:2 and y.=l or v+=1 and v_=2, for which we have ﬁnd/2p. + 14]”2 = 4.899; p. = ,._ = 2, for which we have V3r’2u?2[u_ + Ml“? = 16.0000; and v+:2 and v, =3 or v+'=3 and v- :2, for which we have usl’2v3r'2[y_ + mm = 32.86. -|W“’="1J‘:‘“‘7 'l i i 340 Section 2 Solutions and Comments on Problems 9.23 A quantity of 5 X 10"1 mol kg" of trirnethylammonium chloride [(CH3)3NHCll is added to a water soiution at 298.15 K containing 0.2 mol kg’1 of NaCl. Compute the H30+ concentration in the solution. Assume complete ionization of NaCl and (CH3)3NHC1. K, for (CH3)3N is 6.25 x 10—10at 298.15 K. Solution The important reactions are (CH3}3NHC1 -l- xI-IIO ——> (CH3)3NH(+W + Cltgq], NaCl + xHZO —> Nag,“ + Clygqy (CH3)3,NH(;,U + H20 —> [-130qu + (CHQJN, and H20 + H20 : H3O(:q, + 011;,” The problem states that we may assume that the first two reactions go to completion. Since we have added 5 X 10’4mol kg" of (CH3)3NHC1, we get 5 X 10‘4 mol kg"1 of (CH3)3NH(’;q) and ClL'a'q). We also get 0.2 mol kg! of Naijq) and Clgq) from ionization of the NaCl. The hydrolysis constant for the third reaction is given by Eq. 9.92 for strong conjugate acids: K... L 1.0 x 10'14 K = 7 — z 1.6 >< 10-10. 1 h K, 6.25X10'5 H We see that CMK,l = 8.0 X 10'”. Therefore, we do not have CHKJ >> K... Consequently, we must consider the effect of the self-ionization of water on the H3O+ concentration. The presence of the NaCl means we cannot set the mean ionic activity coefficient for the ionization of water equal to unity. The ionic strength of the solution is given by i=K S = 0.5 2 6,2? i=1 : 0.5[(0.2}(1)2 + (0.2)(1)2 + 5 x 10_4(1)2 + 5 x 10 4(1):] : 0.2005, (2) since the concentrations of Nafan and (:1qu are each 0.20 mol kg’1 and there are 5 X 10'4 mol kg’1 of (CH3)3NH(:qy and Chg,“ present from the ionization of the (CH3)3NHCI salt. At this concentration, we need to use the Davies empiri- cal forrnula to estimate 7/1. Equation 9.68 is 112 lOgW.) *0.50926z+z_]: - 0.305] 1 + S”2 (02005)“2 —0.50926(1)(1) ———e. .1 + (0.2005?!2 “ 0.30(0.2005)] = €0.1268‘7. (3) Thus, 'yi = 107012637 2 0.7467 and Y: = 0.5576. For the water ionization equilibrium, we have Kw = vim(l-I30+)m(OH’), (4) which we can write in the form Kw 1.0 x 10‘H ' -r—=K’,,= HO+ ' =4———= . ’H. y: l m( 3 )m(OH ) 0.5576 1793 X 10 (5) The development leading up to Eq. .997 shows that, for a strong conjugate acid, we expect to have CoKh Kw . "10130? = = [Kw + C.,r'<;,]1-'2 (6) _ + —— [Kw + Com-"2 [Kw + om”? Chapter 9 Thermodynamics of Electrolytic Solutions 341 when the contribution from the self—ionization of H20 is considered. The same result will hold for this problem, wherein we must include the effect of 1/1, pro— vided that we replace K m in Eq. 6 with K ;,,. This gives m(H30+) = [1.793 >‘< 10*14 + (5 x 10‘4)(1.6 >< 10—10))”: : .13 x 10‘7molkg“. (7) 9.24 A quantity of 0.2 mol kg1 of trimethylammonjum hypochlorite [(CH3)3NHCIO] is added to water at 298.15 K. COmpute the H30* concentration in the solution. Assume complete ionization of the (CH3)3NHClO salt. K, for (CH3)3N is 6.25 X 10010 at 298.15 K, and K, for hypochlorous acid (HCIO) is 3.0 X 10'8 at 298.15 K. Solution There are four reactions that need to be considered: (A) (CH3)3NHC10 + xHZO —> (CH3)3NH(‘;q).+ C105,”, (B) (CH3)3NH(:q) + H20 = H3O(;qj + (CH3)3N, l (C) Clog,“ + H20 = HClO + OHfaqj, and (D) H20 + H10 = H3O(§q) + OHQq). The hydrolysis constant for the strong conjugate acid in reaction (B) is K... 2 1.0 x 10’H K“ = — w h " K, 6.25 x 1.0—5 = 1.6 X 10710 (1) The hydrolysis constant for the strong conjugate base in reaction (C) is K 1.0 x 10*“ K"=-—‘”=—=3.33><10*7. 2 '1 K, 3.0 x 10-B ( ) The development given in the text leading up to Eq. 9.100G shows that if we set the mean ionic activity coefﬁcient for the self-ionization of water to unity and ignore the effect on C U of reactions (B) and (C), the H30+ concentra- tion will be given by l“'” H “l COK‘i‘, CaKf, CnKi + Kw ' However, the ionic strength here is substantial, since We have 0.20 mol_kg" of both (CH3)3NH(:qJ and ClOfaq). This gives an ionic strength of ”1011300 = CuKi (3) s = 0.5 5222.2 = 0.5[(0.2)(1)2 + (0.2)(1)2] = 0.20. (4) i=1 As a result, we cannot ignore the effect of 7‘s in reaction (D). The Davies empir- ical formula, Eq. 9.68, gives ' 1/2 log(‘y.-) = —0.50926z,z.[1 + 51/2 — 0.305] ‘ *0 509260) Di (MW 030 02 ] 012681 ' -.( 1+(0.2}1/2 ' ( ' } _ ' - (5) Thus, 'yi : 10‘0‘12'381 = 0.7467? and "y: = 0.55766. Therefore, for the water- ionization equilibrium, we have K. = vim(H30*.)m(0H‘ ). (a) ...
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