# 查看完整版本 : volumetric analysis

cyw1226 2013-8-21 09:25 PM

## volumetric analysis

in an experiment, 2.0 M sodium hydroxide solution was added to 20.0 cm^3 of 1.0 M sulphuric acid until the acid was just completely neutralized. What is the concentration of sodium sulphate in the resultant solution?

chungkin81 2013-8-21 11:52 PM

In an experiment, 2.0 M sodium hydroxide solution was added to 20.0 cm^3 of 1.0 M sulphuric acid until the acid was just completely neutralized. What is the concentration of sodium sulphate in the resultant solution?

2NaOH (aq) + H2SO4 (aq) --> Na2SO4 (aq) + 2H2O (l)
No. of moles of H2SO4 = 1 x 20 / 1000 = 0.02 mol
For just completely neutralized, No. of moles of NaOH = 2 x 0.02 = 0.04 mol
Let the volume of NaOH(aq) required be V,
2 x V / 1000 = 0.04
V = 20 cm^3
Total volume of solution = 20 + 20 = 40 cm^3
No. of moles of Na2SO4 formed = 0.02 mol
Concentration of Na2SO4 = 0.02 / (40 / 1000) = 0.5 mol dm-3

thomas1126 2013-8-22 12:26 AM

2NaOH + H2SO4 -> Na2SO4 + 2 H2O

20.0 cm^3 of 1.0 M sulphuric acid 要用 20.0 cm^3 of 2.0 M sodium hydroxide solution

cyw1226 2013-8-22 07:47 PM

[quote]原帖由 [i]thomas1126[/i] 於 2013-8-22 12:26 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=369955052&ptid=22339811][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
2NaOH + H2SO4 -> Na2SO4 + 2 H2O

20.0 cm^3 of 1.0 M sulphuric acid 要用 20.0 cm^3 of 2.0 M sodium hydroxide solution

:smile_35:

cyw1226 2013-8-22 07:47 PM