Post a New Question


posted by on .

A 25.00-mL sample of an H2SO4 (2 and 4 are subscripted) solution of unknown concentration is titrated with a .1328 M KOH solution. A volume of 38.33 mL of KOH was required to reach the endpoint. What is the concentration of the unknown H2SO4 (again, the 2 and 4 are subscripted) solution?

my work:
38.33 mL KOH* (1L/1000mL)*(.1328 M KOH/L KOH)*(1 mol H2SO4/1mol KOH)=.005090 mol H2SO4

M= mol solution/L= .005090mol/.025L

However, the answer in the book says it should equal .1018M. I'm confused.

  • Chemistry - ,

    I'm sorry... I accidently reposted this--it is not answered yet though.

  • Chemistry - ,

    look at you last ( ). If you balance the reaction, you will find it takes 2mol KOH to neutralize one mole of H2SO4

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question