The neutralization of 27.3 mL of an unknown concentration H2SO4 solution requires 72.2 mL of 0.12 M LiOH solution. What is the concentration of the H2SO4 solution (in M)?

see the last post you did.

To find the concentration of the H2SO4 solution, we need to use the concept of stoichiometry and the equation of the reaction.

The balanced chemical equation for the reaction between H2SO4 and LiOH is:

H2SO4 + 2LiOH -> Li2SO4 + 2H2O

From the equation, we can see that the ratio of H2SO4 to LiOH is 1:2.

Given that 72.2 mL of 0.12 M LiOH solution is required to neutralize the H2SO4, we can use this information to find the moles of LiOH used.

Moles of LiOH = volume (L) × concentration (M) = 72.2 mL × 0.12 M × (1 L / 1000 mL) = 0.00864 moles

Since the H2SO4 to LiOH ratio is 1:2, the moles of H2SO4 used will be half that of LiOH.

Moles of H2SO4 = 0.00864 moles / 2 = 0.00432 moles

Now, we need to find the concentration of the H2SO4 solution. To do this, we divide the moles of H2SO4 by the volume in liters:

Concentration (M) = Moles / Volume (L) = 0.00432 moles / (27.3 mL × 1 L / 1000 mL) = 0.158 M

Therefore, the concentration of the H2SO4 solution is 0.158 M.