H2SO4 + 2KOH -> K2SO4 + 2H2O
From the balanced chemical equation, we can see that 1 mole of H2SO4 reacts with 2 moles of KOH. Therefore, the number of moles of KOH required to neutralize 25.0 mL of 1.48 M H2SO4 is:
Moles of H2SO4 = (25.0 mL / 1000 mL/L) * 1.48 M = 0.037 moles H2SO4
Moles of KOH required = 2 * 0.037 moles H2SO4 = 0.074 moles KOH
The molarity of the KOH solution can be calculated using the formula:
Molarity = Moles of solute / Volume of solution in liters
Volume of KOH solution used = 27.29 mL = 0.02729 L
Molarity of KOH = 0.074 moles KOH / 0.02729 L = 2.71 M
Therefore, the molarity of the KOH solution is 2.71 M.
