A volume of 50.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (HSO4). What was the molarity of the KOH solution if 24.7 mL of 1.50 M H2O4 was needed? The equation is
2KOH(aq) + H2O4(aq) →K2SO4(aq) + 2H20(I)
To determine the molarity of the KOH solution, we first need to use the balanced chemical equation to find the mole ratio between KOH and H2SO4.
From the balanced equation, we can see that 2 moles of KOH react with 1 mole of H2SO4. Therefore, the molarity of KOH can be calculated using the formula:
Molarity of KOH = (Molarity of H2SO4) x (Volume of H2SO4) / (Volume of KOH)
Molarity of KOH = (1.50 mol/L) x (24.7 mL) / 50.0 mL
Molarity of KOH = (1.50) x (24.7) / 50
Molarity of KOH = 0.735 mol/L
Therefore, the molarity of the KOH solution is 0.735 M.