A solution of .12 mol/L of phosphoric acid completely neutralizes .20L of a solution of sodium hydroxide with.a concentration of .15mol/L What is the neutralizing volume of the phosphoric acid?
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.
To find the neutralizing volume of the phosphoric acid, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between phosphoric acid (H₃PO₄) and sodium hydroxide (NaOH).
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.