A volume of 60.0mL of aqueous potassium hydroxide (KOH ) was titrated against a standard solution of sulfuric acid (H 2 SO 4 ). What was the molarity of the KOH solution if 25.7mL of 1.50 M H 2 SO 4 was needed? The equation is

2KOH(aq)+H 2 SO 4 (aq)�¨K 2 SO 4 (aq)+2H 2 O(l)

you need twice the moles of KOH as H2SO4

molesKOH=2*molesH2SO4
molaritiy*.0257L=2*1.50*.0257

solve for molarity.

should be 3 right?

To find the molarity of the KOH solution, we'll use the concept of stoichiometry and the equation provided. Here's how you can calculate it:

1. Start by writing the balanced equation for the reaction:
2 KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2 H2O(l)

2. Use the stoichiometry of the balanced equation to determine the ratio between the reactants and products. In this case, for every 2 moles of KOH, we need 1 mole of H2SO4.

3. Calculate the number of moles of H2SO4 used in the reaction:
Moles of H2SO4 = Molarity × Volume
Moles of H2SO4 = 1.50 M × 0.0257 L (convert the volume to liters)

4. Using the stoichiometry of the balanced equation, we find that for every 2 moles of KOH, we need 1 mole of H2SO4. Therefore, the number of moles of KOH used in the reaction is half the moles of H2SO4.

5. Calculate the number of moles of KOH:
Moles of KOH = Moles of H2SO4 / 2

6. Finally, calculate the molarity of the KOH solution:
Molarity of KOH = Moles of KOH / Volume (in liters)

By following these steps and plugging in the values from your question, you should be able to solve for the molarity of the KOH solution.