How much H3PO4 is needed to neutralize 1 liter of a 0.75M solution of NaOH? (in moles)
To determine the amount of H3PO4 needed to neutralize the 0.75M NaOH solution, we need to consider the balanced chemical equation for the neutralization reaction between H3PO4 and NaOH.
The balanced equation is:
H3PO4 + 3NaOH → Na3PO4 + 3H2O
From the balanced equation, we can see that 1 mole of H3PO4 reacts with 3 moles of NaOH.
Given that the NaOH solution has a concentration of 0.75M, we can conclude that there are 0.75 moles of NaOH in 1 liter of the solution.
Since the ratio of H3PO4 to NaOH is 1:3, we would need 0.75/3 = 0.25 moles of H3PO4 to neutralize 1 liter of the NaOH solution.
To determine the amount of H3PO4 needed to neutralize the NaOH solution, you need to use the balanced chemical equation for the reaction between H3PO4 and NaOH:
H3PO4 + 3NaOH → Na3PO4 + 3H2O
From the equation, you can see that it requires 1 mole of H3PO4 to react with 3 moles of NaOH.
First, calculate the number of moles of NaOH present in the 1 liter of solution:
Moles of NaOH = Molarity × Volume
= 0.75 mol/L × 1 L
= 0.75 mol
Next, use the stoichiometry of the balanced equation to determine the number of moles of H3PO4 required:
Moles of H3PO4 = (Moles of NaOH) ÷ 3
= 0.75 mol ÷ 3
= 0.25 mol
Therefore, you would need 0.25 moles of H3PO4 to neutralize 1 liter of the 0.75M solution of NaOH.