To determine the volumes of 250 mL and 750 mL NaOH solutions needed to prepare 1 L of NaOH solution that can be neutralized by 10 mL of 0.5 M H₃PO₄, we first need to understand the stoichiometry of the neutralization reaction between NaOH and H₃PO₄.

Understanding the Reaction

Phosphoric acid (H₃PO₄) is a triprotic acid, meaning it can donate three protons (H⁺ ions). The balanced chemical equation for the reaction between NaOH and H₃PO₄ can be represented as:

• H₃PO₄ + 3 NaOH → Na₃PO₄ + 3 H₂O

This indicates that one mole of H₃PO₄ reacts with three moles of NaOH. Therefore, we need to calculate how many moles of NaOH are required to neutralize the given amount of H₃PO₄.

Calculating Moles of H₃PO₄

First, we calculate the moles of H₃PO₄ in the 10 mL of 0.5 M solution:

• Volume of H₃PO₄ = 10 mL = 0.010 L
• Molarity of H₃PO₄ = 0.5 M
• Moles of H₃PO₄ = Molarity × Volume = 0.5 mol/L × 0.010 L = 0.005 moles

Determining Moles of NaOH Required

Since one mole of H₃PO₄ requires three moles of NaOH, the moles of NaOH needed for 0.005 moles of H₃PO₄ is:

• Moles of NaOH = 3 × Moles of H₃PO₄ = 3 × 0.005 = 0.015 moles

Concentration of the Final NaOH Solution

Next, we need to find the concentration of the NaOH solution that will be prepared. We want to prepare 1 L (1000 mL) of NaOH solution containing 0.015 moles of NaOH:

• Concentration of NaOH = Moles/Volume = 0.015 moles / 1 L = 0.015 M

Calculating the Required Volumes of NaOH Solutions

Now, we need to mix the 250 mL and 750 mL NaOH solutions to achieve this concentration. Let’s denote:

• V₁ = volume of 250 mL solution
• V₂ = volume of 750 mL solution

Let’s assume the concentrations of the 250 mL and 750 mL solutions are C₁ and C₂ respectively. We can express the total moles of NaOH from both solutions as:

• Total moles = (C₁ × V₁) + (C₂ × V₂)

Since we want the total volume to be 1 L, we have:

• V₁ + V₂ = 1000 mL

To find the specific volumes, we can use the fact that the total moles of NaOH must equal 0.015 moles. However, without the exact concentrations of the 250 mL and 750 mL solutions, we can assume they are both at 1 M for simplicity:

• 0.015 = (1 × V₁) + (1 × V₂)

Solving the Equations

From the volume equation, we can express V₂ as:

• V₂ = 1000 - V₁

Substituting this into the moles equation gives:

• 0.015 = V₁ + (1000 - V₁)

Solving this leads to:

• 0.015 = 1000

This indicates that we need to adjust our assumptions about the concentrations. However, if we assume the 750 mL solution is more concentrated, we can set a ratio based on the volumes. If we take 750 mL of the 750 mL solution and 250 mL of the 250 mL solution, we can achieve the desired concentration.

Final Volume Calculation

To find the exact volumes, we can test the options provided:

• Option (1) 750 mL: Too high, as it exceeds 1 L.
• Option (2) 500 mL: This could work if the concentrations are adjusted accordingly.
• Option (3) 250 mL: This would not provide enough NaOH.
• Option (4) None: This is unlikely as we can achieve the target.

After evaluating the options, the most plausible solution is option (2) 500 mL, assuming the concentrations of the solutions are appropriate to achieve the required moles of NaOH.