A solution of 0.314M KOH is used to neutralize 10.0mL H3PO4 solution.

H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(l) If 29.3mL KOH solution is required to reach the endpoint, what is the molarity of the H3PO4 solution?

See your post above.

To find the molarity of the H3PO4 solution, we need to use the balanced chemical equation and the given information.

According to the balanced chemical equation:
1 mole of H3PO4 reacts with 3 moles of KOH.

Given:
Volume of KOH solution used = 29.3 mL = 0.0293 L
Molarity of KOH solution = 0.314 M

To determine the moles of KOH used, we can use the equation:

moles of KOH = molarity × volume in liters
moles of KOH = 0.314 M × 0.0293 L = 0.00918 moles of KOH

Since the stoichiometric coefficient of KOH is 3, the ratio of moles of KOH to moles of H3PO4 is 1:3. Therefore, the moles of H3PO4 present in the solution are:

moles of H3PO4 = 3 × 0.00918 = 0.0275 moles of H3PO4

Now, we can calculate the molarity of the H3PO4 solution. The molarity is defined as moles of solute per liter of solution:

Molarity = moles of solute / volume of solution in liters
Molarity of H3PO4 = 0.0275 moles / 0.010 L (since 10.0 mL = 0.010 L) = 2.75 M

Therefore, the molarity of the H3PO4 solution is 2.75 M.