To find the molarity of sulfuric acid, H2SO4, it is titrated with .75M KOH. It requires 328.5 ml of KOH to neutralize a 40ml sample of sulfuric acid. Calculate the concentration (mol/L) of the sulfuric acid.
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H2SO4+2KOH>>>K2SO4+2H2O.
Mole ratio= 1:2
vol of acid in ltres= 328.5/1000=
0.3285.
Vol of base in ltrs= 40/1000=0.040
formular
conc of acid x vol of acid acidratio
---------------------------- = ---------
concof base x volofbase baseratio
so: CA x 0.3285 1
------------- = ---
0.75 x 0.040 2
then:.. CA= 0.03/0.657 = 0.04566mol/litre.
Conc of sulphuric acid is 0.406m/ltr.
Mole ratio= 1:2
vol of acid in ltres= 328.5/1000=
0.3285.
Vol of base in ltrs= 40/1000=0.040
formular
conc of acid x vol of acid acidratio
---------------------------- = ---------
concof base x volofbase baseratio
so: CA x 0.3285 1
------------- = ---
0.75 x 0.040 2
then:.. CA= 0.03/0.657 = 0.04566mol/litre.
Conc of sulphuric acid is 0.406m/ltr.
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This answer is incorrect:
Mol KOH in 328.5mL of 0.75M KOH = 328.5/1000*0.75 = 0.246 mol KOH
This will react with 0.123 mol H2SO4
40mL H2SO4 contains 0.123mol
1000mL contains 1000/40*0.123 = 3.075M
Mol KOH in 328.5mL of 0.75M KOH = 328.5/1000*0.75 = 0.246 mol KOH
This will react with 0.123 mol H2SO4
40mL H2SO4 contains 0.123mol
1000mL contains 1000/40*0.123 = 3.075M
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Each OH- -ion of the KOH will react with each H+ -ion of the sukfuric acid. Calculate, how many OH- ions are used to neutralize the acid. You get, how much H+ was in the acid and the acid's volume is known. Take into account how much H is in the sulfiric acid molecules.