- Mass of hydrochloric acid = 73g
- Volume of hydrochloric acid solution = 1 litre


- Volume of hydrochloric acid solution required for neutralization of sodium hydroxide obtained by allowing 0.46g of metallic sodium to act upon water


When metallic sodium reacts with water, it forms sodium hydroxide (NaOH) and hydrogen gas (H2). The balanced chemical equation for this reaction is:

2Na + 2H2O -> 2NaOH + H2

From the equation, we can see that 2 moles of sodium (Na) react with 2 moles of water (H2O) to produce 2 moles of sodium hydroxide (NaOH) and 1 mole of hydrogen gas (H2).

To calculate the volume of hydrochloric acid solution required for neutralization, we need to determine the number of moles of sodium hydroxide produced from the reaction of metallic sodium with water.


1. Determine the number of moles of sodium using the molar mass of sodium (Na):

Molar mass of Na = 23g/mol
Number of moles of Na = Mass of Na / Molar mass of Na
Number of moles of Na = 0.46g / 23g/mol
Number of moles of Na = 0.02 mol

2. From the balanced chemical equation, we know that 2 moles of sodium react to produce 2 moles of sodium hydroxide. Therefore, the number of moles of sodium hydroxide produced is also 0.02 mol.

3. To neutralize sodium hydroxide, an equal amount of hydrochloric acid (HCl) is required. From the balanced chemical equation, we know that 2 moles of sodium hydroxide react with 2 moles of hydrochloric acid. Therefore, the number of moles of hydrochloric acid required is also 0.02 mol.

4. Now, we can calculate the volume of hydrochloric acid solution required using the concentration of the hydrochloric acid solution:

Volume of hydrochloric acid solution = Number of moles of hydrochloric acid / Concentration of hydrochloric acid solution

Given that the concentration of the hydrochloric acid solution is 73g/L, which means there are 73g of hydrochloric acid in 1 litre of solution.

Molar mass of HCl = 36.5g/mol
Number of moles of HCl = 0.02 mol
Concentration of hydrochloric acid solution = 73g/L

Volume of hydrochloric acid solution = 0.02 mol / (73g/L / 36.5g/mol)
Volume of hydrochloric acid solution = 0.02 mol / 2 mol/L
Volume of hydrochloric acid solution = 0.01 L or 10 mL

Therefore, a volume of 10 mL of the hydrochloric acid solution containing 73g of acid per litre would be sufficient for the exact neutralization of sodium hydroxide obtained by allowing 0.46g of metallic sodium to act upon water.