Asked by Piyush
85ml of 0.1M h2so4 solution is titrated against 0.1N koh solution . Calculate the concentration of salt formed when the acid is one fourth neutralized.
Answers
Answered by
DrBob222
You have 85 mL x 0.1M H2SO4 = 8.5 mmols H2SO4.
You want to decrease this by 1/4 or 8.5/4 = 2.125 mmols. That will take 2.125 x 2 = 4.25 mmols KOH
.......H2SO4 + 2KOH ==> K2SO4 + 2H2O
I......8.5......0........0.......0
add............4.25...............
C.....-2.125..-4.25......2.125......
E......6.375.....0.......2.125
So you will have 2.125 mmols K2SO4 formed. What volume will you have.
Since M = mmols/mL, then mL KOH = mmols/M = 4.25/0.1 = 42.5 mL of the 0.1 M KOH. Total volume is 85 + 42.5 = ?
M K2SO4 = mmols/mL = ?
Post your work if you get stuck.
You want to decrease this by 1/4 or 8.5/4 = 2.125 mmols. That will take 2.125 x 2 = 4.25 mmols KOH
.......H2SO4 + 2KOH ==> K2SO4 + 2H2O
I......8.5......0........0.......0
add............4.25...............
C.....-2.125..-4.25......2.125......
E......6.375.....0.......2.125
So you will have 2.125 mmols K2SO4 formed. What volume will you have.
Since M = mmols/mL, then mL KOH = mmols/M = 4.25/0.1 = 42.5 mL of the 0.1 M KOH. Total volume is 85 + 42.5 = ?
M K2SO4 = mmols/mL = ?
Post your work if you get stuck.
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