【正文】 +]= molL1 mL/( mL+)= molL1 [NH3]= molL1 mL/( mL+ mL)= molL1 由 Ksp=[Mn2+][OH]2，得 17 [OH] = ]/ [M n 132sp ?? ??KmolL1=106 molL1 由][NH ]][O H[NH 34b ???K， 得 ]OH[ ]NH[]NH[ 16 5b34 ??? ????? ?? ??? KmolL1 m(NH4Cl) = molL1 L gmol1= g 27. 在 L molL1 H3PO4 溶液中 ， 加入 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)？ {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 =]PO[H ]][HPOO[H 42 243 ? ??， 得 [H3O+]= 108 molL1， pH= (2) П =∑icRT =[c(HPO42)+c(H2PO4)+c(Na+)]RT =(++3) molL1 Jmol1K 1(273+37) K 1JL1kPa? =644 kPa (3) 溶液的滲透濃度為 18 gL1/(180 g mol1) + (++3) molL1= molL1， 18 與血液相比，為高滲溶液。 28. 將 500 ml c(MgCl2)= molL1 和 500 ml c(NH3H 2O) = molL1 混合。求： (1)混合后溶液是否有沉淀生成？請通過計(jì)算加以說明。 (2)若有沉淀要加入多少克 NH4Cl，才能使溶液無 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(NH3H 2O) = molL1 ∵ cKb20Kw, c/Kb500 ∴ [OH] = 13153b Lm )NH( ???? ???????cK 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]≤ 1122sp Lm ]Mg[ ??? ???K=106 molL ∴ [NH4+]= 16523b Lm ]OH[ O]HNH[ ???? ?? ?????K= molL1 x= molL1(+) mol1= g Exercises 1. mL of molL1 propanic acid, HPr, is diluted to mL. What will the final pH of the solution be? (Ka=105) Solution c(HPr)= mL mL ol 1 ?? ? = molL1, [H3O+ ] = 15a Lm o ?? ????cK =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, ?323 NHCHCH . Solution pH=, pOH== [OH]=103 molL1 Kb(CH3CH2NH2) = 423223323 )(] NHCH[C H ]OH [] NHCH[C H ??? ????? 11414323a )NHCH(CH ???? ?????K 3. Pivaic acid is a monoprotic weak acid. A molL1 solution of pivalic acid has a pH=. What is 19 the pH of molL1 sodium pivalate at the same temperature? Solution Ka(HPi)= 5233 )(] [H P i ]Pi[] O[H ??? ?????， Kb(NaPi)=Kw/Ka(HPi)=1014/(105)=109 1519bb Lm ol101 .0Lm ol 0 .1 0 0101 .0]OH[ ???? ????????? cK ， 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)= 2?c ＝ (102)2= 105 (2) When pH=， [H3O+]=103 molL1 [HA]＝ 1523a23 Lm )(]OH[ ???? ????K = molL1 5. 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) 74sp )(S rS O ??? KSmolL1=104 molL1 (2) S＝ 1724 4sp Lm ]SO[)(S rS O ?? ???K ＝ 106 molL1 6. In a saturated solution of calcium phosphate, the concentration of ?34PO ion is 107 molL1. Calculate the Ksp of Ca3(PO4)2. Solution Suppose the solubility of Ca3(PO4)2 is S Ca3(PO4)2(s) 3 Ca2+(aq)＋ 2 PO43(aq) 3S 2S Based on the title， [PO43]＝ 107 molL1＝ 2S So S＝ 107 molL1 Ksp＝ [Ca2+]3[PO43]2＝ (3S)3 (2S) 2＝ 332 2(107)5＝ 1032 7. (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? 20 Ksp for PbSO4 = 108, Ag2SO4 =105. Solution (1) The concentration of Pb2+ is molL1, 182 4spP bS O24 Lm ]Pb[ )(P bS O]SO[4??? ???? K =108 molL1 The concentration of Ag+ is molL1, 12 52 42spSOAg24 Lm ]Ag[ )SO(A g]SO[42??? ???? K=104 molL1 Thus, the PbSO4 separates first with a controlled addition of SO42. (2) When [SO42] is 104 molL1, Ag2SO4 is just ready to precipitate, 14 8244sp2 Lm ]SO[ )(P bS O]Pb[ ???? ????? K =105 molL1 染色地放松對反訴