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‰lÀmub„vÅ_‰agöNÿ9hnc¶n¦^ïyĉR ÿ(W¾–¶n5uã‰(¶n²m-N/f Qi > Ksp¡HR0Œ2#4#+çÿ#ª&IðH ðˆ ƒ ð0ƒ“ŽŸ‹”Þ½h¿ÿ ?ð ÿÿÿ€€€Ì™33ÌÌÌÿ²²²îï€ùÿÿÿ ªð¢ð8ð:ð( ð ð8ðð8 ðP§  ð8ðP§ ðö ð8 “ ð6€¤A¸¿ƒ¿ÀÿðP§  ðˆŸ æ10-1-4 T»yP[He”^ŒTÐvHe”^ (Wq™ŒT¶n²mBaSO4-NX[(W NRs^aˆÿ BaSO4(s) Ba2+(aq) + SO42-(aq) T»yP[He”^OBaSO4¶n㉦^M–NO0 ¡0Œ2Z0‚2#   !)çÿ!#+çÿ #+ #+çÿ+#%ÿþ !)çÿ!ªn% ðZB ð8 s ð*D¿ÀÑÿð°Ð ÐðZB ð8B s ð*D¿ÀÑÿð€` ðH ð8 ƒ ð0ƒ“ŽŸ‹”Þ½h¿ÿ ?ð ÿÿÿ€€€Ì™33ÌÌÌÿ²²²î<ï€ùk ÔðÌð<ðdð( ð ð<ð, ð< “ ð6€ÈZ¸¿ƒ¿Àÿð 0  ðÆŸ  ÐvHe”^§Nu„vŸSàV/f¶n²m-N»yP[:_¦^žX'Y €O gHeSm¦^ÿ;m¦^ ÿÏQ\@bó0 T»yP[He”^8^ g4OHe”^ ÿvQÓ~œg/føvÍS„v ÿ;N‰ÖS³Q£Íyq_ÍT'Y ÿù[ŽNz¶n²mïS N€Q†ÐvHe”^0  ReQÇÏ‘„v‰lÀmBRïSO‰lÀm»yP[‰lÀmŒ[hQ0 ÿN,‚Sm¦^\ŽN10-5mol.l-1 ÿ¤‹:N‰lÀmŒ[hQ ÿ ¡|†0x2'ÿþQ#!)!) !#ªq ðH ð< ƒ ð0ƒ“ŽŸ‹”Þ½h¿ÿ ?ð ÿÿÿ€€€Ì™33ÌÌÌÿ²²²îï€ùÿÿÿ ˜ð ð@ð(ð( ð ð@ððð8 ðàé`G ð@ðéà`Gð ð@ “ ð6€ðg¸¿ƒ¿Àÿðàé`  𦟠ì30pHf6qI-HQ_ËY‰lÀm ÿàVƒ[—‰„vAg+Sm¦^\ ÿSm¦^ïyQiHQ'YŽN¶n¦^ïy0 ù[ŽN TN{|‹W ÿ_NïSÔkƒKspegnxš[0¡f7Œ!)2 !)!)çÿ!ªf ¦ððH ð ƒ ð0ƒ“ŽŸ‹”Þ½h¿ÿ ?ð ÿÿÿ€€€Ì™33ÌÌÌÿ²²²î< ï€ùÿÿÿ ÔðÌàð hñðXð( ð ðhð ð8 ðP€À• ð hð€PÀ•ðÜ¢ ðh “ ð6€Àº¿ƒ¿ÀÿˆðP€À• ðnŸ ú‰lÀm„vlS ù[ŽNBaSO40CaSO4Ç‘(u Nð¹eÕlý Ný€¶n㉠ÿïSåN)R(uKspÿCaCO3 ÿ<KspÿCaSO4 ÿ, ReQ³Ï‘„vNa2CO3 ÿhQèlS:NCaCO3ÿdkÇ zëS‰lÀmlS ÿ ÿQ(ux‘¶nã‰0 CaSO4 (s) SO42_ + Ca2+ Na2CO3 2Na+ + CO32_ CaCO3 ¡Ê0‡2[8‡20‡2 #+çÿ#+çÿ#'¥!þ/¥!þçÿ'¥!þ/¥!þçÿ'¥!þ/¥!þçÿ'¥!þ/¥!þçÿ #+çÿ#+çÿ #+çÿ#'ÿþ #   çÿ  çÿ  <  <    <  çÿ  çÿ    çÿ  <  D  çÿ ª† ˜ð`B ðh ƒ ð0D¿ÀÑÿˆðP` ` ð`B ðhB ƒ ð0D¿ÀÑÿˆð  Ð ð`B ðh ƒ ð0D¿ÀÑÿˆðÀ    ðZB ðh s ð*D¿ÀÑÿð`` @P ðZB ð h s ð*D¿ÀÑÿð` @€ ðH ðh ƒ ð0ƒ“ŽŸ‹”Þ½h¿ÿ ?ð ÿÿÿ€€€Ì™33ÌÌÌÿ²²²îøï€ùÿÿÿ ðˆðð xð ð( ð ðxðèðF ðPð… ðx ðˆðPð…ðF ðx “ ð6€À¼º¿ƒ¿ÀÿðPð… ðØŸ ø T{|‹W„v¾–¶nir( ÿ‰lÀmlS„v z¦^ ÿÖS³QŽNƒ[ìN¶n¦^ïy„vøvù['Y\ ÿøvî]'Y ÿlSŒ[hQ0 ‚Y N‹O‰lÀmlSÍS”^:N ÿ CaSO4 + CO32_ CaCO3 + SO42_ Kl ž[E–1u NRÍS”^Ä~bÿ CaSO4 (s) Ca2+ + SO42_ K1=KSP(CaSO4) Ca2+ + CO32_ CaCO3 K2= 1/KSP(CaCO3 ) ¡10s2T 2:0‡2> 2" ' "  "   çÿ  çÿ  <  çÿ  çÿ  <  çÿ <  çÿ    çÿ  < çÿ çÿ çÿ <      çÿ  <  çÿ çÿ çÿ çÿ çÿ ª,E3 wðZB ðx s ð*D¿ÀÑÿð` @ ðZB ðxB s ð*D¿ÀÑÿð0 € €ðZB ðx s ð*D¿ÀÑÿð@ ð ðZB ðxB s ð*D¿ÀÑÿð@ À@ ðZB ðx s ð*D¿ÀÑÿðð pð ðZB ð xB s ð*D¿ÀÑÿð` @ ðH ðx ƒ ð0ƒ“ŽŸ‹”Þ½h¿ÿ ?ð ÿÿÿ€€€Ì™33ÌÌÌÿ²²²îÞï€ùk vðnð|ðð( ð ð|ðÎð8 ð€À€° ð|ðÀ€€°ð ð| “ ð6€à亿ƒ¿Àÿð€À€° ð°Ÿ ”9hncYÍ‘s^aˆÄ‰‹_ÿ ÙÍylS8^8^«ˆºNìN)R(u ÿ‚YÿYt•¢Wÿ+TCaSO4 ÿ ÿ1uŽNCaSO4¾–¶nŽNx‘ ÿ)R(uNa2CO3OKNlS:NCaCO30¡°K‘(#+çÿ#+çÿ #+çÿ#+çÿ #+çÿ#ªH$ð`² ð| ƒ ð0A ƒ¿Àÿ?ð °5 ðH ð| ƒ ð0ƒ“ŽŸ‹”Þ½h¿ÿ ?ð ÿÿÿ€€€Ì™33ÌÌÌÿ²²²î ï€ùÿÿÿ 8ð0ð€ðÈð( ð ð€ð0¢ ð€ ƒ ð0€ø¼¿ƒ¿Àÿð€@9  ðП î gsQ‰lÀmÍS”^„v¡‹—{:y‹O ‹O1 (W25! N ÿ\úVSOAgCl >eeQ¯~4l-N ÿ N­edÌbv^Oû|ß~-N giRYO„v*g¶n㉄vAgCl0àQ)YT ÿnxž[¾0R‰lÀm-¶nã‰s^aˆöe ÿKmš[AgCl „v¶n㉦^:N1.92×10-3 g·L-10Õ‹Blå‹)n¦^ NAgCl „v¶n¦^ïy0¡8x 2 #  ÿþ$C$$C$$C$$C$ $C$$C$$C$$C$$C$$C$$ª†  ðX ð€ ƒ ð0…‡ƒ¿Àÿð P°ðH ð€ ƒ ð0ƒ“ŽŸ‹”Þ½h¿ÿ ?ð ÿÿÿ€€€Ì™33ÌÌÌÿ²²²îŠï€ :ð2 ð ðÊð( ð ð ð’ ð  “ ð6€D¼¿ƒ¿Àÿðð€P ð,Ÿ ã‰ÿò]åwMr (AgCl) = 143.3 ÿ\AgCl „v¶n㉦^USMObc—{:Nmol ·L-1 ÿRvQ¶n㉦^S :Nÿ S( mol·L-1) =1.92 ×10-3/143.3 = 1.34×10-5 mol·L-1 GP¾‹(WAgCl q™ŒT¶n²m-N ÿ¶n㉆N„vAgCl Œ[hQ㉻yÿ AgCl (s) Ag+ (aq) + Cl- (aq) s^aˆSm¦^/(mol·L-1) S S K sp ¸(AgCl) = {c(Ag+)}{c(Cl-)} = {S}2 = 1.80×10-10 T7h ÿïSÎN¶n¦^ïy¡‹—{¾–¶n5uã‰(„v¶n㉦^0¡œ  2CC CKCc c‚ c Cc k c CKCc Cc Cc k C C"CCC# C‚ # CKCCKªT   2   ðH ð  ƒ ð0ƒ“ŽŸ‹”Þ½h¿ÿ ?ð ÿÿÿÿÿÌfÿÿ3fÿÿ3ÿÿî ï€ ¾ ð¶ 0ð¤ðN ð( ð ð¤ð| 𤠓 ð6€ôE¼¿ƒ¿Àÿðð` ðŸ z‹O2 ò]åw25!öeAg2CrO4 „v¶n¦^ïy:N1.1×10-12 ÿ Õ‹BlAg2CrO4 (s)(W4l-N„v¶n㉦^ÿg·L-1 ÿ0¡B> C  C  C K çÿC K çÿC  C  C K  C  C  C K  ª6  ð’ 𤠓 ð6€˜^¼¿ƒ¿Àÿð`ðÀï  ð,Ÿ ã‰ÿ¾‹Ag2CrO4(s)„v¶n㉦^:Nx mol·L- 0 Ag2CrO4(s) 2Ag+ÿaq ÿ+CrO42-(aq) s^aˆSm¦^/(mol·L-1) 2x x K sp ¸ ÿAg2CrO4 ÿ= {c(Ag+)}2 {c(CrO42-)} 1.1×10-12 = 4x3, x = 6.5×10-5 Mr (Ag2CrO4) = 331.7 ÿ Ag2CrO4 (W4l-N„v¶n㉦^S :Nÿ S = 6.5×10-5×331.7 g·L-1 = 2.2×10-2 g·L-1¡h  2AIçÿAIçÿAA‚ AAAIçÿAIAAIçÿIAAA! A‚ ! A‚ AAIçÿI AAIAA‚ AA‚ A AIçÿAIçÿ AAIçÿAIçÿAAAAA AAAª " k*ðH 𤠃 ð0ƒ“ŽŸ‹”Þ½h¿ÿ ?ð ÿÿÿÿÿÌfÿÿ3fÿÿ3ÿÿîï€ ÈðÀ@ð°ðXð( ð ð°ð( ð° “ ð6€ •¼¿ƒ¿ÀÿðPÀ€< ðŸ Þ‹O6 ¡‹—{25! NCaF2 (s) (1)(W4l-N ÿÿ2 ÿ(W0.010 mol·L-1Ca (NO3)2 ¶n²m ÿÿ3 ÿ(W0.010 mol ·L-1 NaF ¶n²m-N„v¶n㉦^ÿmol·L-1 ÿ0Ôkƒ NÍyÅ`µQ N¶n㉦^„vøvù['Y\0¡dp 2ÿþEÿþÿþEÿþÿþ EÿþÿþEÿþÿþ Eÿþÿþ EÿþÿþEÿþÿþ EÿþÿþEÿþÿþEÿþÿþEÿþÿþª\    ðð ð° “ ð6€L¤¼¿ƒ¿Àÿð  ðŠŸ  ÿ1 ÿCaF2(s)(W¯~4l-N„v¶n㉦^:NS1 ÿ¾‹S1 = x mol·L- K sp ¸ (CaF2) = {c (Ca2+)}{c (F-)}2 1.4×10-9 = x (2x)2 = 4x3 x = 7.0×10-4 S1 = c (Ca2+) = 7.0×10-4 mol·L-1¡‘ 2AA A AA! A‚ ! 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