電解質(zhì)溶液習題-閱讀頁
2024-08-24 09:29本頁面
【正文】 它是一種抗瘧藥。已知pKb1=，pKb2=。mol1)=103 molL1,x=104 mol mL， mol解 HB與KOH反應(yīng)，生成KB，由于HB過量，組成HBKB緩沖溶液。L1，[B]== molL1Ka=10621. 105 mol已知37℃時Ksp{Fe(OH)3}=1039，pKw=。試討論在pH＝。L1此血液中的Fe3+離子濃度為：結(jié)果表明，在pH，將有超過99％的Fe3+沉淀。L1的Fe3+離子游離存在于血液中，說明血液中的Fe3+是以結(jié)合態(tài)形式（如形成鐵蛋白）穩(wěn)定存在。 molH2O等體積混合；(2) molL1 NH3 mol解 （1）由于n(HCl)=n(NH3)， molL1的NH4Cl，溶液的[H3O+]=molL1pH=（2）由于n(HAc)=n(NH3)， molL1的NH4Ac，[H3O+] =pH=（3）由于n(HCl)=n(Na2CO3)， molL1的NaHCO3，[H3O+]=pH=23. 計算下列溶液的pH值：(1) molL1 Na3PO4等體積混合；(2) molL1HCl溶液等體積混合。L1 V molL1 V molL1= molL1， pH=（2） HCl(aq) + Na2CO3(aq) = NaHCO3(aq)初始時：V molL1反應(yīng)后： V molL1 溶液中[Na2CO3]= [NaHCO3] =(V/2V) molL1Ka2 =，[H3O+ ]= 1011 mol{Ksp(PbI2)=109，Ksp(PbSO4)=108}解 PbI2 = Pb2+ + 2I Ksp(PbI2) = [Pb2+][I]2=S(2S)2=4S3S=[Pb2+] = molL1 PbSO4 = Pb2+ + SO42 Ksp(PbSO4) = [Pb2+][SO42]= S2S=[Pb2+] =molL125. 假設(shè)溶于水中的Mn(OH)2完全解離，試計算：(1) Mn(OH)2在水中的溶解度(molL1 NaOH溶液中的溶解度{假如Mn(OH) 2在NaOH溶液中不發(fā)生其它變化}；(4) Mn(OH) 2在 mol解 Mn(OH)2(s) = Mn2+(aq) + 2OH(aq) [Mn2+][OH]2=Ksp(1) S(2S)2=Ksp, S=molL1(2) [Mn2+] = S =3 .72105 molL1(3) [OH]= molL1=1011 molL1S=molL126. molL1 NH3L1 mL/( mL+)= molL1 mL/( mL+ mL)= molL1=106 molL1m(NH4Cl) = molmol1= g27. L molL1)？{M(NaOH)=，M(C6H12O6)=}解 (1) 反應(yīng)前，溶液中n(H3PO4)= molmol1)= mol,n(NaOH)n(H3PO4)= mol= mol反應(yīng) H3PO4(aq) + NaOH(aq) = NaH2PO4(aq) + H2O(l)初始時/mol 平衡時/mol = 繼續(xù)反應(yīng) NaH2PO4(aq) + NaOH(aq) = Na2HPO4(aq) + H2O(l)初始時/mol 平衡時/mol = 所以平衡時[Na2HPO4]=[NaH2PO4] = molL1， pH=(2) П =∑icRT =[c(HPO42)+c(H2PO4)+c(Na+)]RT=(++3) molmol1L1/(180 g L1= mol28. 將500 ml c(MgCl2)= molH2O) = mol求：(1)混合后溶液是否有沉淀生成？請通過計算加以說明。解 (1)混合后溶液中，c(MgCl2)= molH2O) = molL∴ [NH4+]== molL1(+)L1 propanic acid, HPr, is diluted to mL. What will the final pH of the solution be? (Ka=105)Solution c(HPr)== molL1，pH=2. Ethylamine, CH3CH2NH2, has a strong, pungent odor similar to that ammonia. Like ammonia, it is a base. A molL1Kb(CH3CH2NH2) = 3. Pivaic acid is a monoprotic weak acid. A 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[HA]＝= molL1 with respect to sulfate ion?Solution (1)molL1(2) S＝＝106 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 molL1Ksp＝[Ca2+]3[PO43]2＝(3S)3 (2S) 2＝3322(107)5＝10327. (1) A solution is 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 molL1The concentration of Ag+ is molL1Thus, the PbSO4 separates first with a controlled addition of SO42. (2) When [SO42] is 104 molL119
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