Answer :
The molarity of the LiOH solution required to neutralize 25 ml of the H2SO4 solution is approximately 0.143 M.
To find the molarity of the LiOH solution, we can use the concept of stoichiometry and the equation for the neutralization reaction between LiOH and H2SO4.
The balanced equation for the reaction is:
2 LiOH + H2SO4 → Li2SO4 + 2 H2O
From the equation, we can see that the mole ratio between LiOH and H2SO4 is 2:1. This means that 2 moles of LiOH are required to neutralize 1 mole of H2SO4.
Given that the H2SO4 solution has a molarity of 0.112 M, we can calculate the number of moles of H2SO4 in 25 ml of this solution:
Moles of H2SO4 = Molarity × Volume (in liters)
= 0.112 mol/L × 0.025 L
= 0.0028 moles
Since the mole ratio between LiOH and H2SO4 is 2:1, we can determine the moles of LiOH required:
Moles of LiOH = 2 × Moles of H2SO4
= 2 × 0.0028 moles
= 0.0056 moles
Now, we need to calculate the molarity of the LiOH solution. We are given that it is required to neutralize 39.1 ml of the H2SO4 solution.
Molarity = Moles of LiOH / Volume (in liters)
= 0.0056 moles / 0.0391 L
= 0.143 M
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