If 50 mL of 0.5 M oxalic acid is required to neutralize 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is _________ g

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Solution

Q.
If 50 mL of 0.5 M oxalic acid is required to neutralize 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is _________ g. [2024]

Ans.
(4) $H_{2} C_{2} O_{4} + 2 NaOH \to {Na}_{2} C_{2} O_{4} + 2 H_{2} O$

  By stoichiometry:

   $\frac{n_{NaOH}}{2} = \frac{n_{H_{2} C_{2} O_{4}}}{1}$

   $\frac{M_{NaOH} \times V_{NaOH}}{2} = \frac{M_{H_{2} C_{2} O_{4}} \times V_{H_{2} C_{2} O_{4}}}{1}$

   $\frac{M_{NaOH} \times 25 mL}{2} = \frac{0.5 M \times 50 mL}{1}$

   $M_{NaOH} = 2 M$

For 50 mL, 2M NaOH solution:

   $n_{NaOH} = molarity (M) \times volume (V)$

   $= 2 M \times 50 mL = 100 mmol = 0.1 mol$

   $Mass of NaOH = moles of NaOH \times molar mass of NaOH$

   $= 0.1 mol \times 40 g m o l^{- 1} = 4 g$

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