H3PO4 + 3NaOH ==> 3H2O + Na3PO4
mols NaOH = M x L = ?
Convert mols NaOH to mols H3PO4.
M H3PO4 = mols/L. You have M and mols, solve for L.
mols NaOH = M x L = ?
Convert mols NaOH to mols H3PO4.
M H3PO4 = mols/L. You have M and mols, solve for L.
The balanced chemical equation for the reaction is given as:
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
From the balanced equation, we can see that 1 mole of phosphoric acid (H₃PO₄) reacts with 3 moles of sodium hydroxide (NaOH). This means that the molar ratio between H₃PO₄ and NaOH is 1:3.
Given that the concentration of the sodium hydroxide solution is 0.15 mol/L and the volume is 0.20 L, we can determine the number of moles of NaOH using the formula:
moles = concentration x volume
moles of NaOH = 0.15 mol/L * 0.20 L = 0.03 moles
Since the molar ratio between H₃PO₄ and NaOH is 1:3, we can determine the number of moles of H₃PO₄ required to neutralize the NaOH:
moles of H₃PO₄ = (moles of NaOH) / (molar ratio)
moles of H₃PO₄ = 0.03 moles / 3 = 0.01 moles
Now, let's determine the volume of the phosphoric acid solution required to contain 0.01 moles of H₃PO₄. Given that the concentration of the phosphoric acid solution is 0.12 mol/L, we can use the formula:
volume = moles / concentration
volume of H₃PO₄ = 0.01 moles / 0.12 mol/L = 0.083 L or 83 mL
Therefore, the neutralizing volume of the phosphoric acid is 83 mL.