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½â£ºcH2O=¦ÑH2OMH2O?997.094-3=5.54?10(mol.m) -318.0?10¦«m(H2O)=¦ÊH2OcH2O5.5?10-6-112-1==9.93?10(S.m.mol) 45.54?10-2-22-1?+?-¦«?(HO)=¦Ë(H)+¦Ë(OH)m2mm =(3.498+1.98)?10=5.48?10(S.m.mol)¦«m(H2O)9.93?10-11-9¦Á=?==1.81?10 ¦«m(H2O)5.48?10-2
K=aH+aOH-=a¦Èw2H+=(¦ÃH+cH+c¦È)2=(¦Ác2)¦Èc1.81?10-9?5.54?1042 =()=1.00?10-141000
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½â£º(1) k (NH3¡¤H2O)= k(KCl)¡ÁR (KCl) / R (NH3¡¤H2O) = 3.65¡Á10-2 S¡¤m-1
?m(NH3¡¤H2O)= k(NH3¡¤H2O)/c
=3.6510-4 S¡¤m2¡¤mol-1
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²é±íµÃ£º?m(NH3¡¤H2O)=¦Ëm(NH4) +¦Ëm(OH-)
?= 2.714¡Á10-2 S.m2.mol-1
? = ?m /?m = 0.01345
(2) R (H2O) = R (KCl) k(KCl) / k(H2O) = 3.705¡Á105 ?
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