To determine the volume of a 0.1500 M solution of KOH required to titrate 40.00 ml of a 0.0656 M solution of H3PO4, you can use the concept of stoichiometry and the balanced chemical equation for the reaction between KOH (potassium hydroxide) and H3PO4 (phosphoric acid). The balanced equation for this neutralization reaction is:
H3PO4 + 3KOH → K3PO4 + 3H2O
From the balanced equation, you can see that 1 mole of H3PO4 reacts with 3 moles of KOH.
First, calculate the number of moles of H3PO4 in the 40.00 ml of 0.0656 M solution:
moles of H3PO4 = (0.0656 M) x (0.04000 L) = 0.002624 moles
Now, since the mole ratio between H3PO4 and KOH is 1:3, you'll need three times the moles of KOH to neutralize the H3PO4:
moles of KOH = 3 x moles of H3PO4 = 3 x 0.002624 moles = 0.007872 moles
Now, you can calculate the volume (in liters) of the 0.1500 M KOH solution required:
Volume (L) = moles of KOH / concentration of KOH
Volume (L) = 0.007872 moles / 0.1500 M = 0.05248 L
To express this volume in milliliters, multiply by 1000:
Volume (mL) = 0.05248 L x 1000 mL/L = 52.48 mL
So, you would need 52.48 milliliters of the 0.1500 M KOH solution to titrate 40.00 ml of the 0.0656 M H3PO4 solution.