計算酸鹼值?

Question

原文出題
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.

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

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.

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

Answer 1

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!!

Answer 2

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Answer 3

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=1306030402394 for solution...

2006-03-17 11:02:43 補充：
007 增長智慧乎?

Answer 4

　　上述的幾種酸，其解離方程式如下：　　　HA ----> H+ + A-　Ka = [A-][H+]/[HA] 　　　HB ----> H+ + B-　Ka = [B-][H+]/[HB]　　　HC ----> H+ + C-　Ka = [C-][H+]/[HC] 　　　HD ----> H+ + D-　Ka = [D-][H+]/[HD] 　　　HE ----> H+ + E-　Ka = [E-][H+]/[HE]　　　HF ----> H+ + F-　Ka = [F-][H+]/[HF] 　　　HG ----> H+ + G-　Ka = [G-][H+]/[HG]　　　HJ ----> H+ + J-　Ka = [J-][H+]/[HJ] 　　將上述的方程式代入 Ka 及濃度得：　　　　　[A-][H+] = 1.44x10-4 ---------- (1)　　　　　[B-][H+] = 2.38x10-7 ---------- (2)　　　　　[C-][H+] = 3.06x10-4　　　　　[D-][H+] = 4.2x10-4　　　　　[E-][H+] = 5.75x10-4　　　　　[G-][H+] = 2.38x10-4　　　　　[J-][H+] = 3.2x10-4　　由 (2)/(1) 得　[B-]/[A-] = 1.65x10-3　　　　　　　[B-] = 1.65x10-3[A-]　　同理：　　[C-] = 2.13[A-]　　　　　　　　[D-] = 2.92[A-]　　　　　　　　[E-] = 3.99[A-]　　　　　　　　[G-] = 1.65[A-]　　　　　　　　[J-] = 2.22[A-]　　　且由 (1) 得　[H+] = 1.44x10-4/[A-]　　又 [H+] = [A-]+[B-]+[C-]+[D-]+[E-]+[G-]+[J-]　　　　　　= 12.91[A-]　　故　　　12.91[A-]2 = 1.44x10-4　　 　　　　　　　　　[A-] = 3.34x10-3　　　　　　[H+] = 12.91[A-] = 0.043 M　　　　　　因此，此溶液之 pH 值為 1.37。