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pengjian20: ½ð±Ò+1 2012-09-07 18:06:05
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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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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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xiaoshuchong

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