00 ml of 0.425 m h2so4 solution is required to completely neutralize 23.9 ml of koh solution. what is the molarity of the koh solution

00 mL? I believe that you have erroneously pasted the question here.

Anyway, let us continue computing for the molarity of KOH but i will just use a variable "A" for the volume of H2SO4 solution.

The complete balanced reaction is:

H2SO4 + 2KOH - - > K2SO4 + 2H2O

We see that 2 moles of KOH is needed for every mole of H2SO4. So calculate the moles of H2SO4.

moles H2SO4 = 0.245 M * (A / 1000) = 2.45x10^-4 A

So the number of moles of KOH is:

moles KOH = 2.45x10^-4 A * 2

moles KOH = 4.90x10^-4 A

So the molarity of KOH is:

Molarity KOH = 4.90x10^-4 A / (0.0239 L)

Molarity KOH = 0.0205 * A

Simply multiply 0.0205 by the correct volume of H2SO4 in ml