To solve the problem step by step, we need to find the volume of 0.2 M NaOH solution that should be mixed with 500 ml of 0.5 M NaOH solution so that 300 ml of the final solution can be completely neutralized by 20 ml of 2 M H₃PO₄ solution.
### Step 1: Calculate the moles of H₃PO₄
First, we need to calculate the number of moles of H₃PO₄ that will react with the NaOH. The reaction between NaOH and H₃PO₄ can be represented as follows:
\[ \text{H}_3\text{PO}_4 + 3 \text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + 3 \text{H}_2\text{O} \]
From the equation, we can see that 1 mole of H₃PO₄ reacts with 3 moles of NaOH.
The moles of H₃PO₄ in 20 ml of 2 M solution can be calculated as:
\[
\text{Moles of H}_3\text{PO}_4 = \text{Molarity} \times \text{Volume} = 2 \, \text{mol/L} \times 0.020 \, \text{L} = 0.04 \, \text{mol}
\]
### Step 2: Calculate the moles of NaOH required
Since 1 mole of H₃PO₄ requires 3 moles of NaOH, the total moles of NaOH required for 0.04 moles of H₃PO₄ is:
\[
\text{Moles of NaOH required} = 3 \times \text{Moles of H}_3\text{PO}_4 = 3 \times 0.04 = 0.12 \, \text{mol}
\]
### Step 3: Calculate the final concentration of NaOH in the solution
We need to find the final concentration of NaOH in the 300 ml of the final solution. Let \( V_1 \) be the volume of 0.2 M NaOH solution to be added. The total volume of the final solution will be:
\[
V_{\text{final}} = 500 \, \text{ml} + V_1
\]
### Step 4: Set up the equation for moles of NaOH
The total moles of NaOH from both solutions can be expressed as:
\[
\text{Moles from 0.5 M NaOH} + \text{Moles from 0.2 M NaOH} = 0.12 \, \text{mol}
\]
Calculating the moles from each solution:
\[
\text{Moles from 0.5 M NaOH} = 0.5 \, \text{mol/L} \times 0.500 \, \text{L} = 0.25 \, \text{mol}
\]
\[
\text{Moles from 0.2 M NaOH} = 0.2 \, \text{mol/L} \times \frac{V_1}{1000} \, \text{L} = 0.0002 \, V_1 \, \text{mol}
\]
Setting up the equation:
\[
0.25 + 0.0002 \, V_1 = 0.12
\]
### Step 5: Solve for \( V_1 \)
Rearranging the equation gives:
\[
0.0002 \, V_1 = 0.12 - 0.25
\]
\[
0.0002 \, V_1 = -0.13
\]
This indicates an error in the approach, as the moles cannot be negative. We need to re-evaluate our calculations.
### Step 6: Correct the equations
We need to ensure that the total moles of NaOH equals the moles required for neutralization:
\[
0.25 + 0.0002 V_1 = 0.12
\]
Rearranging gives:
\[
0.0002 V_1 = 0.12 - 0.25
\]
\[
0.0002 V_1 = -0.13
\]
This indicates that the volume of NaOH needed is not feasible with the given concentrations.
### Final Calculation
After correcting the approach, we find that:
\[
0.2 V_1 + 250 = 0.4 \times (500 + V_1)
\]
Solving this equation yields:
\[
0.2 V_1 + 250 = 200 + 0.4 V_1
\]
Rearranging gives:
\[
250 - 200 = 0.4 V_1 - 0.2 V_1
\]
\[
50 = 0.2 V_1
\]
\[
V_1 = \frac{50}{0.2} = 250 \, \text{ml}
\]
### Final Answer
The volume of 0.2 M NaOH solution that should be mixed is **250 ml**.
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