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?

Dude I dunno lmao

22.2M

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

Let's start by calculating the moles of KOH used from the volume and concentration provided. Given that the volume of the 0.314 M KOH solution used is 29.3 mL, we can convert this to liters:

29.3 mL = 29.3 mL * (1 L / 1000 mL) = 0.0293 L

Next, we can calculate the moles of KOH:

moles of KOH = volume of solution (in L) * molarity of KOH
= 0.0293 L * 0.314 M
= 0.0091872 moles KOH

From the balanced chemical equation, we see that 1 mole of H3PO4 reacts with 3 moles of KOH. Therefore, the moles of H3PO4 can be calculated:

moles of H3PO4 = (moles of KOH) / 3
= 0.0091872 moles KOH / 3
= 0.0030624 moles H3PO4

Finally, we calculate the molarity of the H3PO4 solution using the volume of the H3PO4 solution that was neutralized:

moles of H3PO4 = molarity of H3PO4 * volume of H3PO4 solution (in L)
0.0030624 moles H3PO4 = molarity of H3PO4 * 10.0 mL * (1 L / 1000 mL)
molarity of H3PO4 = 0.0030624 moles H3PO4 / (10.0 mL * (1 L / 1000 mL))
= 0.30624 M

Therefore, the molarity of the H3PO4 solution is 0.30624 M.