A volume of 40.0 of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the solution if 20.2 of 1.50M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)--->K2SO4(aq)+2H2O(l)

You need units on those volumes.

To determine the molarity of the aqueous potassium hydroxide (KOH) solution, we need to use the balanced equation and the information provided.

The balanced equation for the reaction is:
2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)

From the equation, we can see that it takes 2 moles of KOH to react with 1 mole of H2SO4 to form 1 mole of K2SO4 and 2 moles of H2O.

Given that 20.2 mL of 1.50 M H2SO4 was needed to titrate the 40.0 mL of KOH solution, we can use the equation to find the number of moles of H2SO4 used.

1. First, convert the volume of H2SO4 from milliliters (mL) to liters (L):
20.2 mL = 20.2 / 1000 L = 0.0202 L

2. Calculate the number of moles of H2SO4 used:
moles H2SO4 = volume (L) × molarity (mol/L)
moles H2SO4 = 0.0202 L × 1.50 mol/L = 0.0303 mol

3. Since the stoichiometry of the balanced equation is a 1:2 ratio between H2SO4 and KOH, the number of moles of KOH will be twice that of H2SO4:
moles KOH = 2 × moles H2SO4 = 2 × 0.0303 mol = 0.0606 mol

4. Finally, calculate the molarity of KOH by dividing the moles of KOH by the volume in liters:
molarity of KOH = moles KOH / volume (L)
molarity of KOH = 0.0606 mol / 0.040 L = 1.515 M

Therefore, the molarity of the aqueous potassium hydroxide (KOH) solution is approximately 1.515 M.