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fds329: ½ð±Ò+2, Ó¦ÖúÖ¸Êý+1, ¸ÐлӦÖú¡£ÐÁ¿àÁËO(¡É_¡É)O 2012-09-30 09:50:32
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xiaoshuchong

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2Â¥: Originally posted by xiaoshuchong at 2012-09-06 22:52:40
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3Â¥2012-09-06 22:54:39
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xiaoshuchong

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2Â¥: Originally posted by xiaoshuchong at 2012-09-06 22:52:40
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4Â¥2012-09-06 22:56:03
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xiaoshuchong

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5Â¥2012-09-06 22:58:14
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jiagle

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pengjian20: ½ð±Ò+1 2012-09-07 18:06:05
fds329: ½ð±Ò+3, ¸ÐлӦÖú¡£ÐÁ¿àÁËO(¡É_¡É)O 2012-09-30 09:50:46
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  c(Na+) = (0.5 * 0.010 + 0.5 * 2 * 0.015)/0.200 = 0.10 mol/L
c(H2PO4-) = 0.5*0.010/0.200 = 0.025mol/L
c(HPO42-) = 0.5 * 0.015/0.200 = 0.0375mol/L
pH = pKa - lg[c(H2PO4-) /c(HPO42-) ]
     = 7.199 - 0.176 = 7.02
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6Â¥2012-09-07 15:14:20
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pengjian20

ľ³æ (СÓÐÃûÆø)

ÒýÓûØÌû:
6Â¥: Originally posted by jiagle at 2012-09-07 15:14:20
Õâ¸öÎÊÌâͨ¹ý¼ÆËã¾ÍºÜ·½±ãµÃ³öÁË£º
  c(Na+) = (0.5 * 0.010 + 0.5 * 2 * 0.015)/0.200 = 0.10 mol/L
c(H2PO4-) = 0.5*0.010/0.200 = 0.025mol/L
c(HPO42-) = 0.5 * 0.015/0.200 = 0.0375mol/L
pH = pKa - l ...

lg[c(H2PO4-) /c(HPO42-) ]=lg[0.025/0.0375]=-0.176
PH=pKa+0.176=7.199+0.176=7.375
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7Â¥2012-09-07 17:02:00
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jiagle

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7Â¥: Originally posted by pengjian20 at 2012-09-07 17:02:00
lg=lg=-0.176
PH=pKa+0.176=7.199+0.176=7.375
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8Â¥2012-09-07 18:28:58
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pengjian20

ľ³æ (СÓÐÃûÆø)

ÒýÓûØÌû:
8Â¥: Originally posted by jiagle at 2012-09-07 18:28:58
¼Ó¼õ·ûºÅŪ´íÁË£¬¼ÆËã½á¹ûûÓÐ´í¡£...

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9Â¥2012-09-08 09:14:56
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fds329: ½ð±Ò+1, ¸ÐлӦÖú¡£ÐÁ¿àÁËO(¡É_¡É)O 2012-09-30 09:51:02
ÒýÓûØÌû:
9Â¥: Originally posted by pengjian20 at 2012-09-08 09:14:56
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