? asked in 科學化學 · 1 decade ago

計算酸鹼值?

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.

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.

4 Answers

Rating
  • 1 decade ago
    Favorite Answer

    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

    推薦您前往黃金俱樂部遊戲網!

    提供您詳盡的說明

    您可以先進行遊戲下載:

    網址:http://99點5888th點net

    誠心服務!會員溫馨!永久服務!

    第一:全程直播遊戲的公平與公正性【亞洲唯一直屬會員代理】

    第二:點數儲值和託售皆在十分鐘之內完成【全台娛樂城速度最快24小時不分周末】

    第三:二十四小時專業客服務線上服務全年無休【大大小小問題都可以馬上做詳細解答喔】

    電子機台.真人遊戲和你一同玩樂

    麻將 百家樂 5PK 德州撲克 水果盤 賓果

    輪盤 21點 10點半 骰盅 牌九 鬥地主 13支

    全年網站優惠活動不間斷!

    快去註冊會員!免費:http://99點5888th點net

  • 1 decade ago

    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
    1 decade ago

      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。

Still have questions? Get your answers by asking now.