電解質(zhì)溶液習(xí)題-資料下載頁(yè)
2025-08-05 09:29本頁(yè)面
【正文】 H固體，完全溶解后，設(shè)溶液體積不變，求(1) 溶液的pH值；(2) 37℃時(shí)溶液的滲透壓；(3) 在溶液中加入18 g葡萄糖，其溶液的滲透濃度為多少？是否與血液等滲(300 mmolL1)？{M(NaOH)=，M(C6H12O6)=}解 (1) 反應(yīng)前，溶液中n(H3PO4)= molL1 L= mol,n(NaOH)= g/(40 gmol1)= mol,n(NaOH)n(H3PO4)= mol= mol反應(yīng) H3PO4(aq) + NaOH(aq) = NaH2PO4(aq) + H2O(l)初始時(shí)/mol 平衡時(shí)/mol = 繼續(xù)反應(yīng) NaH2PO4(aq) + NaOH(aq) = Na2HPO4(aq) + H2O(l)初始時(shí)/mol 平衡時(shí)/mol = 所以平衡時(shí)[Na2HPO4]=[NaH2PO4] = molL1由Ka2 =，得 [H3O+]= 108 molL1， pH=(2) П =∑icRT =[c(HPO42)+c(H2PO4)+c(Na+)]RT=(++3) molL1 Jmol1K1(273+37) K=644 kPa(3) 溶液的滲透濃度為18 gL1/(180 g mol1) + (++3) molL1= molL1，與血液相比，為高滲溶液。28. 將500 ml c(MgCl2)= molL1和500 ml c(NH3H2O) = molL1混合。求：(1)混合后溶液是否有沉淀生成？請(qǐng)通過(guò)計(jì)算加以說(shuō)明。(2)若有沉淀要加入多少克NH4Cl，才能使溶液無(wú)Mg(OH)2沉淀產(chǎn)生?（忽略加入NH4Cl固體引起的體積變化）已知Ksp{Mg(OH)2}= 1012，Kb(NH3)=105，Mr(NH4Cl)=。解 (1)混合后溶液中，c(MgCl2)= molL1，c(NH3H2O) = molL1∵ cKb20Kw, c/Kb500∴[OH] = IP=[Mg2+][OH]2=(103)2=107＞Ksp{Mg(OH)2}，有沉淀生成(2)要使沉淀溶解，設(shè)需加入x g NH4Cl，則：[OH]≤=106 molL∴ [NH4+]== molL1x= molL1(+) mol1= gExercises1. mL of molL1 propanic acid, HPr, is diluted to mL. What will the final pH of the solution be? (Ka=105)Solution c(HPr)== molL1, [H3O+ ] ==103 molL1，pH=2. Ethylamine, CH3CH2NH2, has a strong, pungent odor similar to that ammonia. Like ammonia, it is a base. A molL1 solution has a pH of . Calculate the Kb for the ethylamine, and find Ka for its conjugate acid, .Solution pH=, pOH== [OH]=103 molL1Kb(CH3CH2NH2) = 3. Pivaic acid is a monoprotic weak acid. A molL1 solution of pivalic acid has a pH=. What is the pH of molL1 sodium pivalate at the same temperature?Solution Ka(HPi)=，Kb(NaPi)=Kw/Ka(HPi)=1014/(105)=109，pH=4. (1) The weak monoprotic acid HA is % dissociated in molL1 is the acidity constant, Ka, of HA?(2) A certain solution of HA has a pH=. What is the concentration of the solution?Solution (1) Ka(HA)=＝(102)2= 105 (2) When pH=，[H3O+]=103 molL1[HA]＝= molL15. Ksp for SrSO4 is 107. (1) Calculate the solubility of SrSO4 in H2O. (2)What would be the solubility of SrSO4 in a solution which is molL1 with respect to sulfate ion?Solution (1)molL1=104 molL1(2) S＝＝106 molL16. In a saturated solution of calcium phosphate, the concentration of ion is 107 molL1. Calculate the Ksp of Ca3(PO4)2.Solution Suppose the solubility of Ca3(PO4)2 is SCa3(PO4)2(s)3 Ca2+(aq)＋2 PO43(aq) 3S 2S Based on the title，[PO43]＝107 molL1＝2SSo S＝107 molL1Ksp＝[Ca2+]3[PO43]2＝(3S)3 (2S) 2＝3322(107)5＝10327. (1) A solution is molL1 in Pb2+ and molL1 in Ag+. If a solid of Na2SO4 is added slowly to this solution, which will precipitates first, PbSO4 or Ag2SO4? (2) The addition of Na2SO4 is continued until the second cation just starts to precipitate as the sulfate. What is the concentration of the first cation at this point? Ksp for PbSO4 = 108, Ag2SO4 =105.Solution (1) The concentration of Pb2+ is molL1, =108 molL1The concentration of Ag+ is molL1,=104 molL1Thus, the PbSO4 separates first with a controlled addition of SO42. (2) When [SO42] is 104 molL1, Ag2SO4 is just ready to precipitate,=105 molL119
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