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)?
3KOH + H3PO4 ==> 3H2O + K3PO4
mols H3PO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols H3PO4 to mols KOH.
M KOH = mols KOH/L KOH. You know M and mols KOH, solve for L KOH.
To find the volume of KOH solution needed to reach the last equivalence point in the titration, we first need to determine the stoichiometry of the reaction between H3PO4 and KOH.
The balanced chemical equation for the reaction is:
3 H3PO4 + 3 KOH → K3PO4 + 3 H2O
From the equation, we can see that it takes three moles of KOH to react with one mole of H3PO4.
Given that the initial concentration of H3PO4 is 0.40 M and the volume is 35 mL, we can calculate the number of moles of H3PO4 present:
moles of H3PO4 = concentration of H3PO4 × volume of H3PO4 solution
= 0.40 M × 0.035 L
= 0.014 moles of H3PO4
Since the stoichiometry of the reaction is 3:1 (KOH:H3PO4), we know that we will need three times the moles of KOH to neutralize all three acidic protons in H3PO4.
moles of KOH needed = 3 × moles of H3PO4
= 3 × 0.014 moles
= 0.042 moles of KOH
Now, let's find the volume of the 0.30 M KOH solution needed to reach 0.042 moles of KOH:
volume of KOH solution = moles of KOH ÷ concentration of KOH
= 0.042 moles ÷ 0.30 M
= 0.14 L
= 140 mL
Therefore, 140 mL of the 0.30 M KOH solution is needed to reach the last equivalence point in the titration.