In a titration of 35 mL of 0.40 M H3PO4 with 0.30 M KOH solution, what volume (in mL) of KOH solution is needed to reach the last equivalence point (i.e., point in the titration where enough KOH has been added to neutralize all three of the acidic protons in the phosphoric acid)?
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To find the volume of KOH solution needed to reach the last equivalence point, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between H3PO4 and KOH.
The balanced chemical equation for this reaction is:
H3PO4 + 3 KOH -> K3PO4 + 3 H2O
From the balanced equation, we can see that it takes three moles of KOH to react with one mole of H3PO4.
Given:
Volume of H3PO4 solution (VH3PO4) = 35 mL
Molarity of H3PO4 solution (MH3PO4) = 0.40 M
Molarity of KOH solution (MKOH) = 0.30 M
First, let's calculate the number of moles of H3PO4 present in the given volume (VH3PO4).
Number of moles of H3PO4 = (Volume of H3PO4 solution in liters) x (Molarity of H3PO4 solution)
= (35 mL / 1000 mL/L) x (0.40 mol/L)
= 0.014 mol
Since 1 mole of H3PO4 reacts with 3 moles of KOH, we can determine the number of moles of KOH required for the reaction.
Number of moles of KOH = 3 x (Number of moles of H3PO4)
= 3 x 0.014 mol
= 0.042 mol
Now, using the molarity and the number of moles of KOH, we can calculate the volume of KOH solution needed.
Volume of KOH solution = (Number of moles of KOH) / (Molarity of KOH solution)
= 0.042 mol / 0.30 mol/L
= 0.14 L
= 140 mL
Therefore, 140 mL of KOH solution is needed to reach the last equivalence point.