# 計算酸鹼值?

A wastewater solution contains new named acids (i.e., HA, HB, HC, HD, HE, HG, HJ) 0.12 M HA(Ka=1.2E-3), 0.14 M HB (Ka=1.7E-6), 0.17 M HC (Ka=1.8E-3), 0.20 M HD(Ka=2.1E-3), 0.25 M HE(Ka=2.3E-3), 0.17 M HG (Ka=1.4E-3) and 0.20 M HJ(Ka=1.6E-3). Use formulation of charge balance and material balance equations to determine the pH value for this solution. In addition, list the input and output details about how to solve this question via Excel.

Update:

Definition of HA through HJ: (e.g., HA=H+ + A- , thus Ka=[A-][H+]/[HA])

Update 2:

Formal ways to show material balances for [A-]...[G-] are [A-]=[HA]0/(1+[H+]/Ka)=0.12/(1+[H+]/1.2E-3) and so forth. Please try formal procedures in Analytical Chemistry as mentioned before.

Update 3:

Show how to solve it via Excel, since this is a computer problem and incomplet and/or non-formal solution is not accepted.

Rating

大師 請參考本人另一方面答案 就會有收穫 加油勒

2006-03-14 16:36:07 補充：

MB:

(1) 0.12=[HA]+[A-]=[A-](1+[H+]/1.2E-3)

(2) 0.14=[HB]+[B-]=[B-](1+[H+]/1.7E-6)

(3) 0.17=[HC]+[C-]=[C-](1+[H+]/1.8E-3)

(4) 0.20=[HD]+[D-]=[D-](1+[H+]/2.1E-3)

(5) 0.25=[HE]+[E-]=[E-](1+[H+]/2.3E-3)

(6) 0.17=[HG]+[G-]=[G-](1+[H+]/1.4E-3)

(7) 0.20=[HJ]+[J-]=[J-](1+[H+]/1.6E-3)

CB:

[H+]=[A-]+[B-]+[C-]+[D-]+[E-]+[G-]+[J-]+[OH-]

=0.12/(1+[H+]/1.2E-3)+ 0.14/(1+[H+]/1.7E-6)+ 0.17/(1+[H+]/1.8E-3)+ 0.20/(1+[H+]/2.1E-3)+ 0.25/(1+[H+]/2.3E-3)+ 0.17/(1+[H+]/1.4E-3)+ 0.20/(1+[H+]/1.6E-3)+1E-14/[H+]

Using Excel for calculation, one may obtain pH=1.358258

(i.e., [H+]=0.043827026 M).

參考資料

Skoog's Analytical Chemistry.

2006-03-15 08:38:38 補充：

I do not wish to see high school solution herein, so I have to do something. Low-level questions and solutions are filled in the Yahoo Knowledge+ So sad!!

• Anonymous
6 years ago

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Use formulation of charge balance and MATERIAL BALANCE equations...

2006-03-14 09:48:31 補充：

plz c http://tw.knowledge.yahoo.com/question/?qid=130603... for solution...

2006-03-17 11:02:43 補充：

007 增長智慧乎?

• Frank
Lv 7