A 15.0 mL sample of 0.250 M H2SO4 is neutralized by 20.5 mL of KOH solution. What is the molarity of the KOH solution?

To find the molarity of the KOH solution, we can use the concept of stoichiometry and the balanced chemical equation of the neutralization reaction.

The balanced chemical equation for the reaction between H2SO4 (sulfuric acid) and KOH (potassium hydroxide) is:

H2SO4 + 2KOH -> K2SO4 + 2H2O

From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of KOH. Therefore, the mole ratio is 1:2.

First, let's calculate the number of moles of H2SO4 in the 15.0 mL sample. We can use the formula:

moles = molarity * volume (in liters)

moles of H2SO4 = 0.250 M * (15.0 mL / 1000 mL/L) = 0.00375 moles

Since the mole ratio is 1:2 between H2SO4 and KOH, we know that there must be twice as many moles of KOH as there are moles of H2SO4.

Therefore, the number of moles of KOH is:

moles of KOH = 2 * moles of H2SO4 = 2 * 0.00375 = 0.0075 moles

Now, we need to calculate the molarity of the KOH solution.

Molarity is defined as moles of solute divided by liters of solution.

The volume of the KOH solution used in the neutralization reaction is 20.5 mL. However, we need to convert this to liters:

volume (in liters) = 20.5 mL / 1000 mL/L = 0.0205 L

Now we can calculate the molarity of the KOH solution:

molarity (KOH) = moles of KOH / volume (in liters)

molarity (KOH) = 0.0075 moles / 0.0205 L ≈ 0.366 M

Therefore, the molarity of the KOH solution is approximately 0.366 M.

H2SO4 + 2KOH ==> 2H2O + K2SO4

mols H2SO4 = M x L = ?
Look at the equation to convert mols H2SO4 to mols KOH. mols KOH = 2x mols H2SO4
MKOH = mols KOH/L KOH. Solve for M KOH.

0.0250 mol H2S04 X (2 mol KOH/1 mol H2SO4)= 0.05 mol KOH

0.05 mol KOH/0.0150 L KOH= 3.33 M KOH