Check: The units (none) are correct. The magnitude of the answer makes physical sense because the concentration of NH4Cl is high; so it took a significant amount of Mg(OH)2 to raise the pH to 9.00. Note that this number disagrees with the accepted value for the Ksp(Mg(OH)2). This is most likely due to errors in the measurements in this experiment. 18.148 Given: 1.0 L of 0.10 M H2CO3 titrated with NaOH to [H+]=3.2×10−11 M (assume no volume change) Find: mass NaOH Other: Ka1(H2CO3)=4.3×10−7, Ka2(H2CO3)=5.6×10−11 Conceptual Plan: Comparing the [H+] to the Kas, it can be seen that the pH is just below the pKa2 so the final solution is a buffer solution where the acid is HCO3− and the base is CO32−. The total NaOH will be the amount needed to get to the first equivalence point and then what is needed to get to final pH. To first equivalence point, L, M CO32−→mol CO32−→mol NaOH M=molL 1:1 ratio then use the Henderson–Hasselbalch equation to calculate the [CO32−]/[HCO3−].pH=pKa+log[base][acid] Then solve for [CO32−] knowing that [CO32−]+[HCO3−] =initial [H2CO3]. Then L, M CO32−→mol CO32−→mol NaOH M=molL 1:1 ratio beyond first equivalence point then add two NaOH moles and finally total mol NaOH→g NaOH.40.00 g NaOH1 mol NaOH Solution: To the first equivalence point, 1.0 LH2CO3×0.10 mol H2CO31 LH2CO3×1 mol HCO31 mol H2CO3×1 mol NaOH1 mol HCO3=0.10 mol NaOH To the final pH, pH=pKa+log[base][acid] so−log[H+]=−logKa2+log[CO32−][HCO3−]→ −log(3.2×10−11)=−log(5.6×10−11)+log[CO32−][HCO3−]→10.49=10.25+log[CO32−][HCO3−] →0.24=log[CO32−][HCO3−]→[CO32−][HCO3−]=10+0.24=1.73_780 and [CO32−] + [HCO3−] =initial [H2CO3]=0.10 M Thus, [HCO3−]=0.10 M − [CO32−] and [CO32−]0.10M−[CO32−]=1.73_780 → [CO32−]=1.73_780(0.10M − [CO32−]) →[CO32−]=0.173_780 M−1.73_780[CO32−]→2.73_780[CO32−]=0.173_780 M→ [CO32−]=0.0634743 M then 1.0 LCO32−×0.0634743 mol CO32−1 L CO32−×1 mol NaOH1 mol CO32−=0.0634743 mol NaOH. So the total mole NaOH=0.10 mol+0.0634743 mol=0.1634743 mol NaOH. Finally, 0.1634743 mol NaOH×40.00 g NaOH1 mol NaOH=6.5 g NaOH. Check: The units (g) are correct. The magnitude of the answer makes physical sense because if all of the acid were fully titrated, 0.20 mol of NaOH (or 8 g) would be required. The pH indicates that the titration is most of the way there. 18.149 a. Given: Au(OH)3 in pure water Find: molar solubility (S) Other: Ksp=5.5×10−46 Conceptual Plan: Use equations derived in Problems 18.19 and 18.20 and solve for S. Then for ionic compound, AmXn, Ksp=[An+]m[Xm−]n=mmnnSm+n check answer for validity. Solution: For Au(OH)3, Ksp=5.5×10−46, A=Au3+, m=1, X=OH−, and n=3; so Ksp=[Au3+][OH−]3=5.5×10−46=33S4. Rearrange to solve for S.S5.5×10-46274=2.1×10−12 M. This answer suggests that the [OH−]=3(2.1×10−12 M)=6.3×10−12 M. This result is lower than what is found in pure water (1.0×10−7 M), so substitute this value for [OH−] and solve for S=[Au3+]. So Ksp=[Au3+] [OH−]3=5.5×10−46=S(1.0×10−7 M)3 and solving for S gives S=5.5×10−25 M. Check: The units (M) are correct. Because Ksp is so small, the autoionization of water must be considered and the solubility is smaller than what is normally anticipated. b. Given: Au(OH)3 in 1.0 M HNO3Find: molar solubility (S)Other: Ksp=5.5×10−46 Conceptual Plan: Because HNO3is a strong acid, it will neutralize the gold(III) hydroxide (through the reaction of H+ with OH+) to form water (the reverse of the autoionization of water equilibrium). Write balanced equations for dissolving the solid and for the neutralization reaction. Add these two reactions to get the desired overall reaction. Using the rules from Chapter 16, multiply the individual reaction Ks to get the overall K for the sum of these reactions. Then M HNO3, K→S. ICE table Solution: Identify the solid as Au(OH)3. Write the individual reactions and add them together. Au(OH)3(s)+3 H+(aq) ¯ Au(OH)3(s) ⇌Au3+(aq)+3 OH−(aq)Ksp=5.5×10−463H+(aq)+3 OH−(aq)⇌3 H2O(l) (1Kw)3=(11.0×10−14)3 Because the overall reaction is the sum of the dissolution reaction and three times the reverse of the autoion- ization of water reaction, the overall reaction ⇌Au3+(aq)+3 H2O(l) K=Ksp(1Kw)3=(5.5×10−46)(11.0×10−14)3=5.5×10−4=[Au3+][H+]3; then because HNO3 is a strong acid, it will completely dissociate to H+ and NO3−. Set up an ICE table: Au(OH)3(s)+3H+(aq) Au3+(aq)+3 H2O(l) ⇌
[H+]
[Au3+]
Initial
1.0
0.00
Change
−3S
+S
Equil
1.0−3S
+S
K=[Au3+][H+]3=5.5×10−4=S(1.0−3S)3. Assume that S is small (3 S<<1.0), so S(1.0−3 S)3=5.5×10−4 M=S(1.0)3=S. Confirm that the assumption is valid. 3(5.5×10−4)1.0×100%=0.017%<<5%, so the assumption is valid. Check: The units (M) are correct. K is much larger than the original Ksp, so the solubility of Au(OH)3 increases over that of pure water. 18.150 Given: excess AgCl in 0.10 M KI Find: [I−] Other: Ksp(AgCl)=1.77×10−10, Ksp(AgI)=8.51×10−17 Conceptual Plan: Because the KI does not generate a common ion with AgCl, use equations derived in Problems 18.19 and 18.20 and solve for S. So S(AgCl)=[Ag+] then use the equation derived in Problem 18.19 to get Ksp(AgI) for ionic compound, AmXn, Ksp=[An+]m[Xm−]n=mmnnSm+nfor ionic compound, AmXn, Ksp =[An+]m [Xm−]n expression. Substitute value for [Ag+] into Ksp(AgI) expression and solve for [I−]. Solution: For AgCl, Ksp (AgCl)=1.77×10−10, A=Ag+, m=1, X=Cl−, and n=1; so Ksp=[Ag+][Cl−]=1.77×10−10=S2. Rearrange to solve for S. S=1.33×10−5 M. Then Ksp (AgI)=8.51×10−17, A=Ag+, m=1, X=I−, and n=1; so Ksp=[Ag+] [I−]=8.51×10−17. Substitute value for [Ag+] into Ksp(AgI) expression and solve for [I−]. 8.51×10−17=(1.33×10−5)[I−], so [ I−]=6.40×10−12 M Check: The units (M) are correct. Because silver ions are generated from the excess AgCl in solution, the iodide ions are converted to AgI and the concentration of remaining iodide ions is controlled by the solubility of the AgCl. 18.151 Given: 1.00 L of 0.100 M MgCO3 Find: volume of 0.100 M Na2CO3 to precipitate 99% of Mg2+ ions Other: Ksp(MgCO3)=6.82×10−6 Conceptual Plan: Because 99% of the Mg2+ ions are to be precipitated, 1% of the ions will be left in solution. (0.01)(0.100 M Mg2+) Let x= required volume (in L). Calculate the amount of CO32− added and the amount of Mg2+ that does not precipitate and remains in solution. Use these to calculate the [Mg2+] and [CO32−]. The solubility product constant (Ksp) is the equilibrium expression for a chemical equation representing the dissolution of an ionic compound. The expression of the solubility product constant of AmXn is Ksp=[An+]m[Xm−]n. Substitute these expressions in this equation to [Mg2+], [CO32−], Ksp→x. for ionic compound, AmXn, Ksp=[An+]m[Xm−]n Solution: Because 99% of the Mg2+ ions are to be precipitated, 1% of the ions will be left in solution, or (0.01)(0.100 M Mg2+)=0.001 M Mg2+. Let x= required volume (in L). The volume of the solution after precipitation is (1.00+x). The amount of CO32− added =(0.100 M)(x L)=0.100x mol CO32−. The amount of Mg2+ that does not precipitate and remains in solution is (0.100 M)(1.00 L)(0.01)=1.00×10−3 mol, and the amount that precipitates =0.099 mol, which is also equal to the amount of CO32− used. The amount of CO32− remaining in solution is (0.10x−0.099). Thus, [Mg2+]=1.00×10−3mol/(1.00+x) L and [CO32−]=(0.10x−0.099) mol/(1.00+x) L. Then Ksp=6.82×10−6, A=Mg2+, m=1, X=CO32−, and n=1; so Ksp=6.82×10−6= [Mg2+][CO32−]=(1.00×10−3)(0.10x−0.099)(1.00+x)2. Rearrange to solve for x. 1.00+2.00x+x2=1.0×104x−9.9×10−56.82×10−6 → 0=x2−12_.6628x+15_.5161. Using quadratic equation, x=1.3_75 L=1.4 L. Check: The units (L) are correct. The necessary concentration is very low, so the volume is fairly large. 18.152 Given: solution with 0.40 M HCN Find: solubility of CuI Other: Ksp(CuI)=1.1×10−12,Kf(Cu(CN)2−)=1×1024 Conceptual Plan: Write balanced equations for the solubility of CuI and reaction with CN− and expressions for Ksp and Kf. Use initial concentrations to set up an ICE table. Because the K is so large, assume that reaction essentially goes to completion. Solve for [I−] at equilibrium. Solution: Write two reactions and combine. CuI(s)⇌Cu+(aq)+I−(aq) with Ksp =[Cu+][I−]=1.1×10−12Cu+(aq)+2CN−(aq)⇌Cu(CN)2−(aq) ¯ with Kf=[Cu(CN)2−][Cu+][CN−]2=1×1024CuI(s)+2CN−(aq)⇌Cu(CN)2−(aq)+I−(aq) with K=KspKf=[Cu+] [I−][Cu(CN)2−][Cu+][CN−]2=(1.1×10−12)(1×1024)K=[Cu(CN)2−][I−][CN−]2=1_.1×1012 Because Ksp=[Cu−][I−]=1.1×10−12 without HCN present, [Cu+]=[I−]=1.1×10−12=1.0_49×10−6 M. Set up an ICE table with initial concentrations. Because K is so large and because initially [CN−]>[I−], the reaction essentially goes to completion; then write equilibrium expression and solve for x. CuI(s)+2 CN−(aq) Cu(CN)2−(aq)+I−(aq) ⇌
[CN−]
[Cu(CN)2−]
[I−]
Initial
0.40
0.00
1.0_49×10−6
Change
≈−0.40
≈1/2(0.40)
≈1/2(0.40)
Equil
0.40−2x
x
1.04¯9×10−6+x
K=[Cu(CN)2−][I−][CN−]2=(x)(1.0_49×10−6+x)(0.40−2x)2=1_.1×1012. Assume that x>>1.0_49×10−6 M, so (x)(x)(0.40−2x)2=1_.1×1012=(x)2(0.40−2x)2→1_.1×1012=x0.40−2x=1_.049×106→ x=0.1_9995. So [I−]=x=0.1_9995 M=0.2 M I−. Because 1.0_48×10−6 M is insignificant compared to x, the assumption is valid. The solubility of CuI is 0.2 M, or one-half the initial concentration of HCN. Check: The units (M) are correct. Because K is so large, the reaction essentially goes to completion and the solubility of CuI is dramatically increased. 18.153 Given: 1.0 L solution with 0.10 M Ba(OH)2 and excess Zn(OH)2 Find: pH Other: Ksp(Zn(OH)2)=3×10−15, Kf(Zn(OH)42−)=2×1015 Conceptual Plan: Because [Ba(OH)2] =0.10 M,[OH−]= 0.20 M. Write balanced equations for the solubility of Zn(OH)2 and reaction with excess OH− and expressions for Ksp and Kf. Use initial concentrations to set up an ICE table. Solve for [OH−] at equilibrium. Then [OH-]→[H3O+]→pH.Kw=[H3O+][OH−] pH=−log[H3O+] Solution: Write two reactions and combine. Zn(OH)2(s)⇌Zn2+(aq)+2 OH−(aq)with Ksp=[Zn2+][OH−]2=3×10−15Zn2+(aq)+ 24OH−(aq)⇌Zn(OH)42−(aq)with Kf=[Zn(OH)42−][Zn2+][OH−]4=2×1015Zn(OH)2(s)+2OH−(aq)⇌Zn(OH)42−(aq) with K=KspKf=[Zn2+][OH¯−]2[Zn(OH)42−][Zn2+][OH−]42=(3×10−15)(2×1015) K=[Zn(OH)42−][OH−]2=6. Set up an ICE table with initial concentration and solve for x. Zn(OH)2(s)+2 OH−(aq) Zn(OH)42−(aq) ⇌
Initial
[OH−]
[Zn(OH)42−]
0.20
0.00
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