A volume of 50.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 24.7 mL of 1.50 M H2SO4 was needed? The equation is
2KOH(aq) + H2SO4(aq) →K2SO4(aq) + 2H2O(I)
First, calculate the number of moles of sulfuric acid used in the reaction:
moles of H2SO4 = (volume of H2SO4 used) x (molarity of H2SO4)
moles of H2SO4 = 24.7 mL x 0.0015 mol/mL
moles of H2SO4 = 0.03705 mol
Since the reaction is 1:2 between KOH and H2SO4, the moles of KOH used in the reaction will be half of the moles of H2SO4:
moles of KOH = 0.03705 mol / 2
moles of KOH = 0.018525 mol
Now, calculate the molarity of KOH solution:
molarity of KOH = moles of KOH / volume of KOH used (in L)
molarity of KOH = 0.018525 mol / 0.050 L
molarity of KOH = 0.3705 M
Therefore, the molarity of the KOH solution is 0.3705 M.