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发表于 21-7-2007 05:55 PM
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原帖由 jiko 于 21-7-2007 16:53 发表
NAOH in excess
[H+]=(25.00-24.95)(0.01)/(25.00+24.95)
    =1.0x10^-4 mol dm-3
[salt]=(24.95)(0.01)/(25.00+24.95)
      = 0.05mol dm-3
since it was a buffer solution,
pH=pKa+log[sa ...
NaOH哪里有excess?
是CH3COOH多余
这个是一个buffer solution来的
CH3COONa的comcentration= 24.95X0.100/49.95
amount of weak acid left= 0.05X0.100/49.95
[H+]=Ka X [acid]/[salt]
=1.8x10^-5 x 0.05/24.95
pH=7.44 |
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发表于 22-7-2007 08:58 PM
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Calculate the pH of the resulting solution when the 24.95cm3of 0.10 mol dm-3 NaOH have been added to 25.0cm3 of 0.10 mol dm-3 CH3COOH.
Ka of ethanoic acid is 5.56 X 10^-10 mol dm-3
可是你的Ka是1.8x10^-5???
[ 本帖最后由 zxteh 于 22-7-2007 09:00 PM 编辑 ] |
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发表于 22-7-2007 09:48 PM
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发表于 23-7-2007 01:53 PM
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原帖由 nikuang04 于 21-7-2007 05:55 PM 发表
NaOH哪里有excess?
是CH3COOH多余
这个是一个buffer solution来的
CH3COONa的comcentration= 24.95X0.100/49.95
amount of weak acid left= 0.05X0.100/49.95
[H+]=Ka X [acid]/[salt]
=1.8x10^-5 ...
仔细一看,发现做法一样,但是写法不一样。。。。。。
如果acid excess pH 应该没可能7.多吧。。。
数位对吗? |
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发表于 23-7-2007 06:28 PM
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原帖由 jiko 于 23-7-2007 13:53 发表
仔细一看,发现做法一样,但是写法不一样。。。。。。
如果acid excess pH 应该没可能7.多吧。。。
数位对吗?
因为是strong alkali和weak acid的关系吧 |
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发表于 23-7-2007 06:56 PM
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原帖由 nikuang04 于 23-7-2007 06:28 PM 发表
因为是strong alkali和weak acid的关系吧
erm...
可是剩下的是weak acid了哦。。。。
你的Ka 是用哪一个呢? |
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发表于 26-7-2007 07:31 PM
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explain what is the properties of the Al? |
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发表于 26-7-2007 10:03 PM
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原帖由 jiko 于 23-7-2007 01:53 PM 发表
仔细一看,发现做法一样,但是写法不一样。。。。。。
如果acid excess pH 应该没可能7.多吧。。。
数位对吗?
你用错Ka了。 题目给的是Kb=5.56x10^-10 mol dm^-3
所以你应该用 pKa x pKb = 14 找出 pKa 或以 Ka x Kb = 10^-14 找出 Ka。
那么你就可以得到 pH 7.44 了 |
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发表于 26-7-2007 10:11 PM
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发表于 26-7-2007 10:28 PM
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原帖由 jiko 于 23-7-2007 01:53 PM 发表
仔细一看,发现做法一样,但是写法不一样。。。。。。
如果acid excess pH 应该没可能7.多吧。。。
数位对吗?
有可能,因为CH3COOH is a weak acid, thus CHCOONa is a strong conjugate base.
CH3COOH 只剩下一点点,所以pH > 7 |
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发表于 27-7-2007 03:08 PM
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explain what is the properties of the Al?
nikuang04 ,huhuxboy and jiko come and solve this question!!!
[ 本帖最后由 zxteh 于 27-7-2007 03:14 PM 编辑 ] |
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发表于 27-7-2007 03:26 PM
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不好意思。。。
我还没有学inorganic,学校还没教。。。 |
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发表于 27-7-2007 10:42 PM
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原帖由 zxteh 于 27-7-2007 03:08 PM 发表
explain what is the properties of the Al?
nikuang04 ,huhuxboy and jiko come and solve this question!!!
resistance corrosion
我只记得一个 |
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发表于 27-7-2007 10:47 PM
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原帖由 prettyboy07 于 26-7-2007 10:03 PM 发表
你用错Ka了。 题目给的是Kb=5.56x10^-10 mol dm^-3
所以你应该用 pKa x pKb = 14 找出 pKa 或以 Ka x Kb = 10^-14 找出 Ka。
那么你就可以得到 pH 7.44 了
谢你哦
看到了 |
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发表于 28-7-2007 04:13 PM
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explain what is properties of group 2 element?
nikuang04 ,huhuxboy and jiko come and solve this question!!! |
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发表于 29-7-2007 01:08 AM
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Group 2 elements - Be Mg Ca Sr Ba Ra
Atomic Radius- Increases down the grp. The proton number remain constant down the grp. Each succesive elements in the grp has an xtra filled shell of electrons, shielding the attraction of the nuclues towards the valence electrons.
Ionisation energy- Decrease down the group.Hence the reactivity of the elements increases down the grp due to the ease of removing the valance electron. as we go down the grp the atomic radius increase, causing the attraction of nuclues towards the outher shll of electrons to be weak, therefore the ionisaton energy decreases. (Ionisation energy- the enerygy needed to remove one mole of electron from one mole of gaseous atom of an element.)
Electronegativity- Decrease down the group.
Grp 2 elements are reducing agent, they readily give up their outermost electrons to form X2+ ions. This is also shown by their -ve std electrode potential.
As we go down the grp, we'll find that the members at the bottom of the grps are stronger reducing agents.
Reactions:
With Oxygen- burn in O2 to form a WHITE SOLID.
With Water- react with water to form Hydroxides.
Solubility of grp 2 sulphates- decrease down the grp.
Solubility og grp 2 hydroxides- increase down the grp.
Thermal stability of grp 2 salts- increase down the grp. |
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发表于 30-7-2007 08:13 PM
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A buffer solution is made by adding 1.0 mol of CH3COOH and 1.0 mol of CH3C))Na to enough water to make 1.0 dm3 of solution.
Calculate the pH of the buffer. (Ka=1.8 X 10^-5 mol dm-3
Calculate the pH of the buffer solution after 0.100 mol of NaOH is added |
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发表于 5-8-2007 12:17 PM
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大家好,我出来报到。我是今年lower six的。来自柔佛
我学校老师真是神速啊,一下子教physical chemistry教玩the periodic table了。现在跳来electrochemistry。根本不知道老师在讲什么
我也来出题:For a daniell cell with standard electrode potential 1.10 volt, will the electrode potential of the cell be the same, increased or decreased when the concentration of Cu2+ is increased? |
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发表于 5-8-2007 08:49 PM
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the emf of the cell will increase, that is more than 1.10volt.
[ 本帖最后由 huhuxboy 于 5-8-2007 08:52 PM 编辑 ] |
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发表于 5-8-2007 09:21 PM
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原帖由 huhuxboy 于 5-8-2007 20:49 发表
the emf of the cell will increase, that is more than 1.10volt.
你很厉害啊
什么问题都会的
你是什么州的?学校的老师教到哪一课了? |
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