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Sm£¨298K£©/Jmol1K-1 131.0 203.3 175.3
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rHm £¨298K£©=rGm £¨298K£© T rSm £¨298K£©
= - 542 kJmol1 298 ¡Á0.0163 kJmol1 = -537.14 kJmol1
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CaCO3£¨s£© = CaO£¨s£© CO2£¨g£©
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c (H) /c <6.9¡Á10-4
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ƽºâŨ¶È/ mol•L-1 0.08 0.84 0.32


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ln K (383K)= = - 4.65
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t1 = 100 min ʱ:£º
c ( HBr) = ( 10.25 ¨C 10.00 )¡Á a / V = 0.25 a / V (mol • L-1)
c (¦Á-äå´ú±ûËá ) = ( 10.00 ¨C 0.25 )¡Á a / V = 9.75 a / V (mol • L-1)
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= 0.05/5.8 10-10 £¾500
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c(H )= •c = ¡Á1.0 mol•L-1=5.4 10-6 mol•L-1
pH =5.27
= = 100=0.011%

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c(OH-)/ c =2.0¡Á10-3
c(NH3)= ¡Ác =0.22 mol•L-1
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= 0.040/4.2 10-7 £¾500
c(H )= •c = ¡Á1.0 mol•L-1
=1.3 10-4 mol•L-1
c(HCO )=1.3 10-4 mol•L-1
c(CO )= K •c =5.6¡Á10-11 mol•L-1
12£®Å«¸¥¿¨ÒòÊÇÒ»ÖÖÒ»ÔªÈõ¼î£¬²âµÃc=0.020 mol•L-1Å«¸¥¿¨ÒòË®ÈÜÒºµÄpH=10.57£¬¼ÆËãÅ«¸¥¿¨ÒòµÄÀë½â³£ÊýK ¡£
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c(OH-)/ c =3.7¡Á10-4
K = =6.8¡Á10-6
13£®¼ÆËã˵Ã÷c(NaAc)=0.10 mol•L-1µÄNaAcË®ÈÜÒºÓëc(Na2CO3)=0.10mol•L-1µÄNa2CO3Ë®ÈÜÒºµÄ¼îÐÔÇ¿Èõ¡£
½â£º NaAcË®ÈÜÒº
= 0.10/5.7 10-10 £¾500
c(OH-)/ c = = =7.5 10-6
pOH =5.12
pH=8.88
Na2CO3Ë®ÈÜÒº
= 0.10/1.8 10-4 £¾500
c(OH-)/ c = = =4.2 10-3
pH =14-pOH =11.62

14£®Ä³Ò»ËáÓêÑùÆ·µÄpH=4.07£¬¼ÙÉè¸ÃÑùÆ·µÄ³É·ÖΪHNO2£¬¼ÆËãHNO2µÄŨ¶È¡£
½â£º c(H )/ c =8.5¡Á10-5
c(HNO2)= ¡Ác = ¡Á1.0 mol•L-1
=1.4¡Á10-5mol•L-1
½â£º c(H )/ c =8.5¡Á10-5
c(HNO2)= ¡Ác £«c(H )= ¡Á1.0 mol•L-1 8.5¡Á10-5
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(1) K •c/ c =6.4¡Á10-5¡Á0.050£¾20 K
¶øc/ c <20K
¡àc(H )/ c = £½
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pH=2.89
(2) pH=pK -lg =4.19-lg =4.49
16£®½«Å¨¶ÈΪ0.020 mol•L-1±½°·(C6H5NH2)ºÍŨ¶ÈΪ0.020 mol•L-1ÏõËáË®ÈÜÒºµÈÌå»ý»ìºÏ£¬¼ÆËãËùµÃÈÜÒºµÄpHÖµ¡£ [K (C6H5NH2)=4.6¡Á10-10]
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c(C6H5NH2)=0.010 mol•L-1
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c(C6H5NH )=0.010 mol•L-1
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pH=3.33

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pH=pK -lg =7.20-lg =6.90
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pH=pK ¡ªlg
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c(Ac-)=0.18 mol•L-1
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