What volume (in $\mathrm{mL}$ ) of $\mathrm{HCl}$ solution containing $73 \mathrm{~g}$ per litre is required…
What volume (in $\mathrm{mL}$ ) of $\mathrm{HCl}$ solution containing $73 \mathrm{~g}$ per litre is required to completely neutralise sodium hydroxide solution, obtained by allowing $0.46 \mathrm{~g}$ of metallic sodium to act upon water?
$30$
$20$
$10$
$40$
Solution
Step 1: Calculate the moles of sodium First,
we need to determine the number of moles of metallic sodium (\(\mathrm{Na}\)) that reacted. The molar mass of sodium is approximately \(23\mathrm{~g/mol}\). \(\text{Moles\ of\ Na}=\frac{\text{Mass\ of\ Na}}{\text{Molar\ mass\ of\ Na}}=\frac{0.46\mathrm{~g}}{23\mathrm{~g/mol}}=0.02\mathrm{~mol}\)
Step 2: Determine the moles of sodium hydroxide produced The reaction of sodium with water produces sodium hydroxide (\(\mathrm{NaOH}\)) and hydrogen gas:\(2\mathrm{Na}+2\mathrm{H}_{\mathrm{2}}\mathrm{O}\rightarrow 2\mathrm{NaOH}+\mathrm{H}_{\mathrm{2}}\)
From the balanced equation, the mole ratio of \(\mathrm{Na}\) to \(\mathrm{NaOH}\) is 2:2, or 1:1. Therefore, the number of moles of \(\mathrm{NaOH}\) produced is equal to the number of moles of \(\mathrm{Na}\) that reacted.\(\text{Moles\ of\ NaOH}=\text{Moles\ of\ Na}=0.02\mathrm{~mol}\)
Step 3: Calculate the moles of HCl required The neutralization reaction between hydrochloric acid (\(\mathrm{HCl}\)) and sodium hydroxide (\(\mathrm{NaOH}\)) is a 1:1 molar ratio:\(\mathrm{HCl}+\mathrm{NaOH}\rightarrow \mathrm{NaCl}+\mathrm{H}_{\mathrm{2}}\mathrm{O}\)To completely neutralize \(0.02\mathrm{~mol}\) of \(\mathrm{NaOH}\), an equal number of moles of \(\mathrm{HCl}\) are required.\(\text{Moles\ of\ HCl}=\text{Moles\ of\ NaOH}=0.02\mathrm{~mol}\)
Step 4: Calculate the volume of HCl solution The concentration of the \(\mathrm{HCl}\) solution is given as \(73\mathrm{~g}\) per litre. We need to find the volume of this solution that contains \(0.02\mathrm{~mol}\) of \(\mathrm{HCl}\). First, calculate the mass of \(0.02\mathrm{~mol}\) of \(\mathrm{HCl}\).
The molar mass of \(\mathrm{HCl}\) is the sum of the molar masses of \(\mathrm{H}\) (\(1\mathrm{~g/mol}\)) and \(\mathrm{Cl}\) (\(35.5\mathrm{~g/mol}\)), which is \(36.5\mathrm{~g/mol}\).\(\text{Mass\ of\ HCl}=\text{Moles\ of\ HCl}\times \text{Molar\ mass\ of\ HCl}=0.02\mathrm{~mol}\times 36.5\mathrm{~g/mol}=0.73\mathrm{~g}\)
Now, use the concentration to find the volume in litres. $\text{Volume\ (L)}=\frac{\text{Mass\ of\ HCl}}{\text{Concentration}}=\frac{0.73\mathrm{~g}}{73\mathrm{~g/L}}=0.01\mathrm{~L}$ Finally, convert the volume from litres to millilitres. $\text{Volume\ (mL)}=0.01\mathrm{~L}\times 1000\mathrm{~mL/L}=10\mathrm{~mL}$ Answer: The volume of $\mathrm{HCl}$ solution required is $10\mathrm{~mL}$