Re: [化學] 我的化學退化了...
※ 引述《beavertail97 (奏音璃)》之銘言:
: 就是阿
: 有一個題目寫說NaH2PO4 1M 請問其PH值....
: 可是我怎麼也算不出來....
: 懇請各位好心人提示我一下吧
: 感恩
那就推一次吧=..="
pH=(pKa1+pKa2)/2
由於是多元子酸
所以你可以接受當 - +
H2A + H →H3A
- 2- +
或 H2A →HA + H
會這樣解離吧
所以當兩式合在一起時
- 2-
2*H2A →HA + H3A
所以他的平衡常數K:
2-
[H3A][HA ]
K = ___________________
- ^2
[H2A ]
2- + 2-
[H3A][HA ] [H3A] [H ][HA ] 1
______________ = _____________* _____________ = _______ * Ka2
- - + - -
[H2A ][H2A ] [H ][H2A ] [H2A ] Ka1
so
2-
Ka2 [H3A][HA ]
K=_____ =________________ --------------(1)
- ^2
Ka1 [H2A ]
Because - 2-
2*H2A →HA + H3A
1
- 2x x x
____________________________________
1-2x x x
2-
so [HA ]=[H3A]-----------------(2)
(2)代入(1)
2- ^2
Ka2 [H3A][HA ] [H3A]
K=_____ =________________ =_____________----------------(3)
- ^2 - ^2
Ka1 [H2A ] [H2A ]
回到最初的公式
+ ^2 + ^2
[H ] [H3A] 變成(3) [H3A] Ka2 [H ]
_____ = _________ → _______________________ = ________ = ________
- - ^2 ^2
Ka1 [H2A ] [H2A ] Ka1 Ka1
+ ^2 ^2 Ka2
so[H ] = Ka1 * ___________ = Ka1 * Ka2
Ka1
_________________________
所以 + √
[H ] = Ka1 * Ka2
轉換成p-function
pH =(pK1+pK2)/2 (爽...打完了=..=+)
由於輸入有困難
建議你先寫到紙上
會比較好看
只不過是公式代來代去=..=+
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