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rosesea1989(´ð·ÇËùÎÊ´ú·¢): ½ð±Ò+5 2014-11-07 20:03:19
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4Â¥2014-11-07 16:34:30
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robertno

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ÒÔÅäÖÃ1L 0.1mol/L pH7.8µÄÁ×ËáÑλº³åÈÜҺΪÀý:
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pH=pKa2-lgc(HPO4^2-) /c(H2PO4^-)
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c(HPO4^2-) = 0..02 mol/L; c(H2PO4-) =0.08 mol/L
m(NaH2PO4) = 0.1¡Á120 =12g
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m(NaOH) = 0.02¡Á40 = 0.8g

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6Â¥2014-11-09 10:24:57
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gubei

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2Â¥2014-11-07 15:30:33
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3Â¥2014-11-07 15:33:03
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99503102

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5Â¥2014-11-07 17:16:06
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Àîսʿ

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7Â¥2014-11-09 12:23:31
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rosesea1989

ľ³æ (СÓÐÃûÆø)

ÒýÓûØÌû:
4Â¥: Originally posted by ³ÈËÔÚº£ÄÏ at 2014-11-07 16:34:30
pH        0.2mol/L NaH2PO4(ml)        0.2mol/L Na2HPO4(ml)
5.7        93.5        6.5
5.8        92.0        8.0
5.9        90.0        10.0
6.0        87.7        12.3
6.1        85.0        15.0
6.2        81.5        18.5
6.3        77.5        22.5
6.4        73.5        26.5
6.5        68.5        31.5
6.6        62.5        37.5
6.7        5 ...

Ç× Ö±½Ó°Ñ0.02»»Ëã³É0.01¾ÍÐÐÊÇÂð  NaH2PO4ºÍNa2HPO4¼ÓÁ¿»¹ÊÇÒ»ÑùµÄÊÇÂð
ÎÒÊÇѧÅÝÒ»Åö¾ÍËé
8Â¥2014-11-10 08:46:20
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rosesea1989

ľ³æ (СÓÐÃûÆø)

ÒýÓûØÌû:
7Â¥: Originally posted by Àîսʿ at 2014-11-09 12:23:31
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9Â¥2014-11-10 08:47:36
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rosesea1989

ľ³æ (СÓÐÃûÆø)

ÒýÓûØÌû:
6Â¥: Originally posted by robertno at 2014-11-09 10:24:57
ÒÔÅäÖÃ1L 0.1mol/L pH7.8µÄÁ×ËáÑλº³åÈÜҺΪÀý:
»º³åÈÜÒº¼ÆË㹫ʽ£º
pH=pKa2-lgc(HPO4^2-) /c(H2PO4^-)
°ÑpH=7.8ºÍpKa2=7.2´úÈ룬4c(H2PO4^-)/c(HPO4^2-)=4/1
Á×ËáÑÎŨ¶ÈΪ0.1mol/L£¬¾ÍÊÇc(H2PO4^-)+c(HPO4 ...

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10Â¥2014-11-10 08:47:56
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