It takes 56 mL of 0.124 M KOH to neutralize 45.4 mL of H2SO4 solution. What is the concentration of the H2SO4?

The formula is C1V1=C2V2 Where Ci is the concentration and vi is the volume

therefore c2= { C1V1}/ C2

{56 ml * 0.124 mol/dm3 }/45.4 ml = 0.153 mol/dm3

but according to the formula the stoichiometry between KOH and H2SO4 is 2:1

Therefore H2SO4 concentration is 0.153*{1/2} =0.076 mol/dm3

write balanced equation: 2KOH + H2SO4 ? K2SO4 + 2H2O 2mol KOH will react with 1mol H2SO4 mol KOH in 130ml a million.00MKOH answer = a hundred thirty/one thousand*a million = 0.13mol KOH mol H2SO4 in 100ml of 0.500 H2SO4 answer = one hundred/one thousand*0.5 = 0.05mol H2SO4 From the balanced equation: 0.05mol H2SO4 will react with 0.10mol KOH there'll be 0.03mol KOH ultimate unreacted. complete quantity = a hundred thirty+one hundred= 230ml concentration of KOH ultimate in answer = 0.03*one thousand/230 = 0.a hundred thirty M KOH For K2SO4: 1mol H2SO4 (proscribing reactant) produced 1mol K2SO4. you have 0.05mol H2SO4, so which you would be able to have 0.05mol K2SO4 dissolved in 230ml answer concentration K2SO4 = 0.05*one thousand/230 = 0.217M K2SO4