50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is _________.

Explanation

To solve the problem, we need to determine the amount of sodium hydroxide (NaOH) present in 50 mL of the sodium hydroxide solution that is neutralized by 50 mL of 0.5 M oxalic acid.

### Step-by-Step Solution:

1. **Determine the moles of oxalic acid used:**
- Molarity (M) = moles/volume (L)
- Given: Volume of oxalic acid = 50 mL = 0.050 L, Molarity of oxalic acid = 0.5 M
- Moles of oxalic acid = Molarity × Volume = 0.5 mol/L × 0.050 L = 0.025 moles

**Hint:** Remember that Molarity is defined as moles of solute per liter of solution.

2. **Identify the stoichiometry of the reaction:**
- Oxalic acid (H2C2O4) is a dibasic acid, meaning it can donate 2 protons (H+).
- The balanced reaction is:
\[ H_2C_2O_4 + 2 NaOH \rightarrow Na_2C_2O_4 + 2 H_2O \]
- This indicates that 1 mole of oxalic acid reacts with 2 moles of NaOH.

**Hint:** Understanding the stoichiometry of the reaction is crucial for determining the relationship between reactants.

3. **Calculate the moles of NaOH required:**
- From the stoichiometry, 1 mole of oxalic acid requires 2 moles of NaOH.
- Therefore, moles of NaOH needed = 2 × moles of oxalic acid = 2 × 0.025 moles = 0.050 moles.

**Hint:** Always check the coefficients in the balanced equation to find the correct mole ratio.

4. **Determine the concentration of NaOH in the solution:**
- We know that the moles of NaOH in 25 mL (0.025 L) of the sodium hydroxide solution is 0.050 moles.
- To find the concentration of NaOH:
\[ \text{Concentration (M)} = \frac{\text{moles}}{\text{volume (L)}} = \frac{0.050 \text{ moles}}{0.025 \text{ L}} = 2 \text{ M} \]

**Hint:** The concentration can be found by dividing the number of moles by the volume in liters.

5. **Calculate the amount of NaOH in 50 mL of the solution:**
- Since we have a concentration of 2 M, we can find the amount of NaOH in 50 mL (0.050 L) of this solution:
\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 2 \text{ mol/L} \times 0.050 \text{ L} = 0.1 \text{ moles} \]

**Hint:** Use the formula for moles again to find the total amount in a different volume.

6. **Convert moles of NaOH to grams:**
- The molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol.
- Therefore, the mass of NaOH in 50 mL = moles × molar mass = 0.1 moles × 40 g/mol = 4 grams.

**Hint:** To convert moles to grams, multiply the number of moles by the molar mass of the substance.

### Final Answer:
The amount of NaOH in 50 mL of the given sodium hydroxide solution is **4 grams**.