35.00mL of 0.250M KOH solution is required to completely neutralize 10.50mL of H2SO4 solution. what is the molarity of the sulfuric acid solution

KOH(aq)+ H2SO4 -------> K2SO4(aq) +H2O(l

2KOH(aq)+ H2SO4 -------> K2SO4(aq) +H2O(l)

Note that I have balanced the equation.
mols KOH = M x L = ?
moles H2SO4 = mols KOH x 1/2
M H2SO4 = mols H2SO4/L H2SO4

To find the molarity of the sulfuric acid solution, we can use the concept of stoichiometry.

Step 1: Write the balanced chemical equation:
KOH(aq) + H2SO4(aq) → K2SO4(aq) + H2O(l)

Step 2: Note the stoichiometric ratio between KOH and H2SO4:
From the balanced equation, we can see that 1 mole of KOH reacts with 1 mole of H2SO4.

Step 3: Determine the number of moles of KOH:
Moles of KOH = Molarity × Volume = 0.250 M × 35.00 mL
= 8.75 mmol (millimoles) or 0.00875 moles

Step 4: Since the stoichiometric ratio between KOH and H2SO4 is 1:1, the number of moles of H2SO4 is the same as the moles of KOH:
Moles of H2SO4 = 0.00875 moles

Step 5: Convert the moles of H2SO4 to molarity:
Molarity of H2SO4 = Moles of H2SO4 / Volume (in liters)
= 0.00875 moles / (10.50 mL / 1000 mL)
= 0.00875 moles / 0.0105 L
≈ 0.833 M

Therefore, the molarity of the sulfuric acid solution is approximately 0.833 M.

To find the molarity of the sulfuric acid (H2SO4) solution, we'll use the concept of stoichiometry. Stoichiometry is the calculation of reactants and products in a chemical reaction.

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

From the balanced equation, we can see that the stoichiometric ratio between KOH and H2SO4 is 2:1. This means that for every 2 moles of KOH, 1 mole of H2SO4 is required for complete neutralization.

We know the volume and molarity of KOH solution used: 35.00 mL of 0.250 M KOH. Using this information, we can calculate the number of moles of KOH used:
moles of KOH = volume of KOH solution (in L) × molarity of KOH solution
moles of KOH = 35.00 mL × 0.250 M / 1000
moles of KOH = 0.00875 mol

Since the stoichiometric ratio between KOH and H2SO4 is 2:1, the number of moles of H2SO4 used is half of the moles of KOH used:
moles of H2SO4 = 0.00875 mol / 2 = 0.004375 mol

Now, we need to calculate the molarity of the sulfuric acid solution. We know the volume of H2SO4 solution used is 10.50 mL.
molarity of H2SO4 solution = moles of H2SO4 / volume of H2SO4 solution (in L)
molarity of H2SO4 solution = 0.004375 mol / 10.50 mL / 1000
molarity of H2SO4 solution = 0.415 M

Therefore, the molarity of the sulfuric acid solution is 0.415 M.