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ÀÛ¼ºÀÚ Uploader : boat7944 ÀÛ¼ºÀÏ Upload Date: 2017-11-20º¯°æÀÏ Update Date: 2017-11-20Á¶È¸¼ö View : 1062

Kb=[BH+][OH-]/[B]

[OH-] = Kb * [B]/[BH+]

pOH = pKb + log( [BH+]/[B] )

pH + pOH = 14

pH = 14 - pKb + log( [B]/[BH+])
[BH+] in formula = [BH]

*** Âü°í¹®Çå[References] ***

pH = 14+log(Kb)+log([B]/[BH])
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