電解質(zhì)溶液習(xí)題(參考版)
2025-08-08 09:29本頁(yè)面
【正文】 L1, Ag2SO4 is just ready to precipitate,=105 molL1,=104 molL1, =108 molL1 in Pb2+ and molL1＝2SSo S＝107 molL16. In a saturated solution of calcium phosphate, the concentration of ion is 107 molL1=104 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 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 solution of pivalic acid has a pH=. What is the pH of molL1 solution has a pH of . Calculate the Kb for the ethylamine, and find Ka for its conjugate acid, .Solution pH=, pOH== [OH]=103 molL1, [H3O+ ] ==103 mol mol1= gExercises1. mL of molL1x= 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 molL1，c(NH3(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)=。L1混合。L1和500 ml c(NH3L1，與血液相比，為高滲溶液。mol1) + (++3) molK1(273+37) K=644 kPa(3) 溶液的滲透濃度為18 gL1 JL1由Ka2 =，得 [H3O+]= 108 molL1 L= mol,n(NaOH)= g/(40 gL1 H3PO4溶液中， g NaOH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 L gL1由，得molL1由Ksp=[Mn2+][OH]2，得[OH] =molL1 [NH3]= mol mL，為了不使Mn(OH)2沉淀形成，需含NH4Cl多少克?解 溶液中，[Mn2+]= molL1 MnCl2溶液中， molL1 =107 molL1(4) [Mn2+] = molL1S=[Mn2+] =molL1， [OH] =2S = 105 molL1= 105 molL1MnCl2溶液中的溶解度。L1)； (2) Mn(OH) 2飽和溶液中的［Mn2+］和［OH］；(3) Mn(OH) molL1=104 molL1= 103 molL1， pH=24. PbI2和PbSO4的Ksp非常接近，兩者飽和溶液中的［Pb2+］是否也非常接近，通過(guò)計(jì)算說(shuō)明。L1= molL1 V molL1 V molL1Ka2 =， [H3O+ ]= 108 molL1溶液中[Na2HPO4] =[NaH2PO4] = (V/2V) molL1反應(yīng)后： V mol解 （1） H3PO4(aq) + Na3PO4(aq) = NaH2PO4(aq) + Na2HPO4(aq)初始時(shí)：V mol mol molL1/2= molL1/2= molL1=106 molL1/2= molL1Na2CO3溶液等體積混合。H2O等體積混合；(3) mol molL1 NH322. 計(jì)算下列溶液的pH值：(1) mol而在生理?xiàng)l件（37℃，pH=）下，1020 mol解 當(dāng)溶液中99％的Fe3+離子沉淀時(shí)，F(xiàn)e3+(aq) + 3OH(aq) = Fe(OH)3(s) pH=(10－11)=在pH＝，[OH]=107 mol計(jì)算使99％的Fe3+離子沉淀時(shí)的pH值。L1。L1， [H3O+] = 106 mol[HB]== mol mL混合， mL，試求此弱酸(HB)的解離平衡常數(shù)。L1, pOH=, pH=20. molL1pKb1=, Kb1=106, ∵c/Kb1500，設(shè)[OH] = x mol解 cb= g/( g已知1 L水中，計(jì)算該飽和溶液的pH值。L1，pH=（3）NaOH過(guò)量, [OH]=pOH = ，pH=19. 喹啉（C20H24N2O2，摩爾質(zhì)量＝ gL1/2= molL1NaOH溶液，則混合溶液的pH值又是多少?解 （1）應(yīng)該加入堿性的NaAc。L1HCl溶液，問(wèn)：(1) 如使pH=，應(yīng)該加入HAc還是NaAc？(2) mol求HX的Ka。L1的BHX溶液，測(cè)得pH=。L1/(105 molL1 ，則[H3O+]嬰兒=105 molL116. 。設(shè)乳酸的濃度為c(HC3H5O3)。解 pH=, [H3O+]=103 molL1pH = 15. 已知乳酸的Ka=104。[H3O+]=molL1乳酸溶液的pH值。L1時(shí)，Ag2CrO4開(kāi)始沉淀，此時(shí)溶液中剩余的Cl濃度為：[Cl]＝＝105 molL1沉淀Cl所需[Ag+]較小，AgCl沉淀先生成。{已知：Ksp(AgCl)＝10Ksp(Ag2CrO4)＝1012}解 AgCl開(kāi)始沉淀時(shí)：[Ag+]AgCl＝＝108 molL1pH=(pKa1+pKa2)= (+)=13. molL1)=1:2，反應(yīng)為：Na3PO4 + 2HCl = NaH2PO4 + 2NaCl生成[NaH2PO4]= molL1pH=(pKa2+pKa3) =（+）= (2) n(Na3PO4)：n(HCl)=(100 mL molL1)=1:2，反應(yīng)為：H3PO4(aq) + 2NaOH(aq) = Na2HPO4(aq) + 2H2O(l)生成[Na2HPO4]= molL1HCl相混合解 (1) n(H3PO4)：n(NaOH)=(100 mL molL1NaOH相混合（2）100 mL、 molL112. 計(jì)算下列溶液的pH值：（1）100 mL、 molL1[]=[Na+]= molL1)
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