✔ 最佳答案
Molar mass of Na2CO3 = 23x2 + 12+ 16x3 = 106 g/mol
No. of moles of Na2CO3 used to make 250 cm³ solution =3.65/106 = 0.0344 mol
Out of the 250 cm³ Na2CO3 solution, only 25 cm³ is usedin titration.
No. of moles of Na2CO3 used in titration = 0.0344 x(25/250) = 0.00344 mol
Reaction in the titration :
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
No. of moles of Na2CO3 used in titration = 0.00344 mol
No. of moles of HCl used in titration = 0.00344 x 2 = 0.00688 mol
Molarity of HCl used = 0.00688 / (20/1000) = 0.344M
(The given answer 0.0344 M is INCORRECT.)
2013-04-01 04:00:55 補充：
The volume of the solution made is 250 cm³, but only 25 cm³ is used in titration.
The fraction of the solution used in titration = 25/250
所製成溶液的體積是 250 cm³，在滴定中只使用了其中的 25 cm³。
用於滴定的溶液的分率 = 25/250
2013-04-01 04:01:28 補充：
The volume of the solution made is 250 cm³, but only 25 cm³ is used in titration.
The fraction of the solution used in titration = 25/250
所製成溶液的體積是 250 cm³，在滴定中只使用了其中的 25 cm³。
用於滴定的溶液的分率 = 25/250
2013-04-02 02:28:04 補充：
你有 case 便拿出來討論吧！
2013-04-02 23:44:52 補充：
Consider the titration :
CH3COOH + NaOH → CH3COONa + H2O
No. of moles of NaOH used in titration = 0.105 x (15.9/1000) = 0.00167 mol
No. of moles of diluted CH3COOH used in titration = 0.00167 molVolume of diluted
2013-04-02 23:45:14 補充：
CH3COOH used in titration = 25 cm³ = 0.025 dm³
Molarity of diluted CH3COOH used in titration = 0.00167/(0.025) = 0.0668 M
Molarity of CH3COOH in the original solution = 0.0668 x (100/10) = 0.668 M
2013-04-02 23:46:06 補充：
Consider the titration :
CH3COOH + NaOH → CH3COONa + H2O
No. of moles of NaOH used in titration = 0.105 x (15.9/1000) = 0.00167 mol
No. of moles of diluted CH3COOH used in titration = 0.00167 mol
2013-04-02 23:46:26 補充：
Volume of diluted CH3COOH used in titration = 25 cm³ = 0.025 dm³
Molarity of diluted CH3COOH used in titration = 0.00167/(0.025) = 0.0668 M
Molarity of CH3COOH in the original solution = 0.0668 x (100/10) = 0.668 M
