By titration, 66.6 mL of aqueous H2SO4 neutralized 11.4 mL of 0.465 M LiOH solution. What was the molarity of the aqueous acid solution?
See your post just above. All of these are done the same way.
1. Write and balance the equation. H2SO4 + 2LiOH ==> Li2SO4 + 2H2O
2. mols = M x L = ?
3. Convert mols of what you have to mols of what you want. In this case it is mols LiOH to mols H2SO4.
4. M os what you want is M = mols/L
Post your work if you have trouble.
To find the molarity of the aqueous acid solution (H2SO4), we can use the concept of stoichiometry in a neutralization reaction. The balanced chemical equation for the reaction between H2SO4 and LiOH is:
H2SO4 + 2LiOH -> Li2SO4 + 2H2O
According to the equation, one mole of H2SO4 reacts with two moles of LiOH to produce one mole of Li2SO4 and two moles of H2O.
In this case, 11.4 mL of 0.465 M LiOH is neutralized by 66.6 mL of the aqueous H2SO4 solution. We need to figure out the number of moles of LiOH used in the reaction.
Number of moles of LiOH = Volume of LiOH (in L) x Molarity of LiOH
= 11.4 mL x (1 L / 1000 mL) x 0.465 M
= 0.005301 moles
Since the stoichiometric ratio between LiOH and H2SO4 is 2:1, the number of moles of H2SO4 used in the reaction is half of the number of moles of LiOH.
Number of moles of H2SO4 = 0.005301 moles / 2
= 0.00265 moles
Now, we can find the molarity of the aqueous H2SO4 solution by dividing the number of moles of H2SO4 by the volume of the solution in liters.
Molarity of H2SO4 = Number of moles of H2SO4 / Volume of H2SO4 solution (in L)
= 0.00265 moles / (66.6 mL x (1 L / 1000 mL))
= 0.039899 M
Therefore, the molarity of the aqueous H2SO4 solution is 0.039899 M.