Problem: Find the amount of sodium hydroxide in 50 ml of the given solution if 50 ml of 0.5M oxalic acid is needed to neutralize 25 ml of sodium hydroxide solution.

Solution:

To solve this problem, we need to use the concept of acid-base titration. In an acid-base titration, we determine the concentration of an unknown solution by adding a known solution of a strong acid or base until the equivalence point is reached.

Given Data:

- Volume of oxalic acid = 50 ml
- Concentration of oxalic acid = 0.5 M
- Volume of sodium hydroxide solution = 25 ml

Step 1: Write the balanced chemical equation for the reaction between oxalic acid and sodium hydroxide.

H2C2O4 (aq) + 2NaOH (aq) → Na2C2O4 (aq) + 2H2O (l)

Step 2: Calculate the amount of oxalic acid used in the titration.

Number of moles of oxalic acid (H2C2O4) = Molarity × Volume (in L)
= 0.5 × 0.05
= 0.025 moles

Step 3: Use the balanced chemical equation to determine the number of moles of sodium hydroxide used in the titration.

From the balanced chemical equation, 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide. Therefore, the number of moles of sodium hydroxide used in the titration is:

Number of moles of NaOH = 2 × Number of moles of H2C2O4
= 2 × 0.025
= 0.05 moles

Step 4: Calculate the concentration of sodium hydroxide in the given solution.

Concentration of NaOH (in M) = Number of moles / Volume (in L)
= 0.05 / 0.05
= 1 M

Answer: Therefore, the amount of sodium hydroxide in 50 ml of the given solution is 1 M.