Organic, Ka Keq, 望大學生指教, 謝

dolphin_ice · 發表於 23/2/2008 09:37 AM

What is the ultimate difference between the following formulae?
When should each of the formulae below be applied?

1.  Ka=[H+][A-]
            [HA]

2. Ka=[A-][H3O+]
               [HA]
Above = Keq x [H2O]
^How?

3. Keq=[A-][H3O+]
            [HA][H2O]

4. In what circumstances does pKa=pH?

[ 本帖最後由 dolphin_ice 於 23/2/2008 10:50 AM 編輯 ]

fish · 發表於 23/2/2008 11:01 AM

原帖由 dolphin_ice 於 23/2/2008 09:37 AM 發表

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What is the ultimate difference between the following formulae?
When should each of the formulae below be applied?

還是改一改wordings好些

1.
Arrhenius acids: HA  <----->  A-  + H+
Arrhenius acids are molecules that produce H+ ion in aqueous solution.

2.
Bronsted-Lowry acids: HA + H2O <------>  A-  + H3O+
Bronsted-Lowry acids are molecules that donates proton(H+) and does not necessarily to happen in aqueous medium.

4.
Handerson-Hasselbalch equation:
pH = pKa + log([base]/[acid])
when [base] = [acid], pH =pKa
You can use pH =pKa when you are at the half-equivalence point of a titration reaction.
Handerson-Hasselbalch equation can be used in any points of titration EXCEPT starting point and after equivalence point (mole of acid = mole of base).

至於3. 嘛....平日我都是用1.  就算了的，反正 H3O+  <---> H+  + H2O....而H2O在平日計的 (dilute solution) equilibrium 中可以neglect....
即 1. 跟 2. 計的answer都是一樣
只想到若H2O(l)的concentration是重要的話，那可能是high concentration solution了

問elven2001吧(推

只覺topic的"Orgainc"有點令我想了一會

Organic 應注重pKa多點....因為Lewis acid / base 不一定要有/用 H

第5800帖 o_O

[ 本帖最後由 fish 於 23/2/2008 11:39 AM 編輯 ]

elven2001 · 發表於 23/2/2008 01:16 PM

嚮中國大陸的理論入面,呢兩條都係屬於酸性弱電解質量度電離平衝的式子

對應的式子,前提的[A-]同[H3O+]的濃度係可以睇成相等,即[H3O+]的二次方,應改寫成

c為HA的初始濃度
Ka=[H3O+]^2    , c-[H3O+]=c,  [H3O+]=開方(Ka*c)
      c-[H3O+]

而限制在於c/Ka的值係大於等於500,即本身酸的電離度係小於5%
因此好明顯條式係有缺陷.....

因此正確的

[H3O+]=[-Ka+開方(Ka^2+4Kac)]    ..............(1)
                        2
pH=-lg[H3O+]...................(2)

代入先係最準確的式子

或pH=pKa+lg[ ( [H3O+]  ) / ( c-[H3O+] ) ]

懶惰者,把lg[ ( [H3O+]  ) / ( c-[H3O+] ) ]  寫為lg[ 1 / c ]亦不為大過

第3條係弱情況下計算的準確度問題

[ 本帖最後由 elven2001 於 23/2/2008 01:30 PM 編輯 ]

dolphin_ice · 發表於 23/2/2008 09:00 PM

原帖由 fish 於 23/2/2008 11:01 AM 發表

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還是改一改wordings好些

1.
Arrhenius acids: HA A- + H+
Arrhenius acids are molecules that produce H+ ion in aqueous solution.

2.
Bronsted-Lowry acids: HA + H2O A- + H3O+
B ...

So basically, I don't need to care about the formula that I'm not familiar with?

fish · 發表於 24/2/2008 12:14 AM

原帖由 dolphin_ice 於 23/2/2008 09:00 PM 發表

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So basically, I don't need to care about the formula that I'm not familiar with?

I would say the 4 most common formulae you will need to use are

Ka=[H+][A-]
[HA]

Kw = Ka‧Kb

pH = -log[H+]

and

pH = pKa + log([conjugate base]/[acid]) <-----used in buffers and titration

when you are dealing with Bronsted-Lowry acids/ bases, which are the most common in junior years.

對了，我改了一下 Handerson-Hasselbalch equation的wordings...以免混淆

dolphin_ice · 發表於 1/3/2008 10:47 AM

Kw = Ka‧Kb
呢條未學過, but maybe it's because I used other sorts of symbols instead. Symbols in formulae do change from places to places.

Ka=conc of acid?
Kb= conc of base?

Kw <--Is that to do with water?

Is this formula the one for using water to calculate the pH of weak bases?

fish · 發表於 1/3/2008 11:55 AM

原帖由 dolphin_ice 於 1/3/2008 10:47 AM 發表

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Kw = Ka‧Kb
呢條未學過, but maybe it's because I used other sorts of symbols instead. Symbols in formulae do change from places to places.

Ka=conc of acid?
Kb= conc of base?

Kw <--Is that to do with water?

Is this formula the one for using water to calculate the pH of weak bases?

Calculations below drop off [H2O] because they are insignificant here.

For the Kw questions:

Kw is actually a constant (Kw = 1.0 * 10^-14 at 25C), namely ion-product constant or dissociation constant of water. 看不到下圖按我

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Kw = Ka‧Kb can be applied to everywhere. For example, you want to find the equilibrium constant for the following reaction:看不到下圖按我

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However, in most references, only the Ka value of ethanoic acid is given. Using Kw = Ka‧Kb allows you do find the answer.看不到下圖按我

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After obtaining Kb, you can determine concentration and whatever related things.

Ka is a equilibriumconstantat constant temperature that can be used to determine the concentration of acid or vice versa.
Equilibrium constants are customarily madeno unit. (I don't know if your teacher wants the unit, but equilibrium constants actually just determine therelative ratioof reactivities(product and reactant).
Butconcentration must has a unit, which is usually "molarity(M)" in chemistry.

Using the above example( ethanoate + water )
the Kbdetermined is 5.6 * 10^-10, if we are given that the initial concentration of ethanoate salt solution is 1.00M and you are required to find the concentration of each species in the solution (except water) at equilibrium.
Calculation: 看不到下圖按我

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significant figures方面，美國的chem 有自己一套的方法，不是次次都3 sig. fig., 看到不是3 sig. fig.的話不要來打偶

[ 本帖最後由 fish 於 1/3/2008 11:58 AM 編輯 ]

dolphin_ice · 發表於 2/3/2008 09:43 AM

Oh I see!
I've learnt this formula before!

Yes...the sig fig in UK and US are different.
Gave me some confusion at the start. (Taiwan uses the US way)

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Organic, Ka Keq, 望大學生指教, 謝

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