Preparing 5N HCl Solution from 37% Concentrated HCl

This problem asks us to determine the volume of concentrated 37% HCl solution required to prepare 1 liter of a 5N HCl solution. To solve this, we need to relate the concentration of the concentrated HCl to its Normality and then use the dilution formula.

Understanding Normality and Equivalent Weight

Normality (N) is a measure of concentration defined as the number of gram equivalent weights of solute per liter of solution. The formula is:

\( \text{Normality (N)} = \frac{\text{Number of gram equivalents of solute}}{\text{Volume of solution in liters}} \)

The number of gram equivalents is given by:

\( \text{Number of gram equivalents} = \frac{\text{Mass of solute (g)}}{\text{Equivalent Weight of solute (g/equivalent)}} \)

For an acid like HCl, the equivalent weight is equal to its molecular weight because it donates one proton (\(\text{H}^+\)).

• Molecular weight of HCl (\(\text{M}_{\text{HCl}}\)) = Atomic weight of H + Atomic weight of Cl
• \(\text{M}_{\text{HCl}} \approx 1.008 \text{ g/mol} + 35.45 \text{ g/mol} = 36.458 \text{ g/mol}\). Often approximated as 36.5 g/mol.
• Equivalent Weight of HCl = 36.5 g/equivalent.

Required Mass of Pure HCl

We need to prepare 1 liter (1000 mL) of a 5N HCl solution. This means the solution must contain 5 equivalents of HCl per liter.

• Required number of equivalents = 5 equivalents/L \(\times\) 1 L = 5 equivalents.
• Required mass of pure HCl = Required equivalents \(\times\) Equivalent Weight of HCl
• Required mass of pure HCl = \(5 \text{ equivalents} \times 36.5 \text{ g/equivalent} = 182.5 \text{ g}\).

So, we need 182.5 grams of pure HCl to prepare 1 liter of 5N solution.

Concentration of the Source HCl Solution (37%)

The concentrated HCl is given as 37%. While percentage concentration usually means weight/weight, in problems like this without a given density, 37% is often interpreted as 37 grams of pure HCl per 100 mL of solution for practical calculation purposes.

• Assuming 37% means 37 g of HCl per 100 mL of solution.
• Concentration in g/L = \(\frac{37 \text{ g}}{100 \text{ mL}} \times \frac{1000 \text{ mL}}{1 \text{ L}} = 370 \text{ g/L}\).

Converting Source Concentration to Normality

Now, we can convert the concentration of the source solution (370 g/L) into Normality.

• Normality (N1) = \(\frac{\text{Concentration (g/L)}}{\text{Equivalent Weight (g/equivalent)}}\)
• N1 = \(\frac{370 \text{ g/L}}{36.5 \text{ g/equivalent}} \approx 10.137 \text{ N}\).

The concentrated 37% HCl solution (interpreted as 370 g/L) has a Normality of approximately 10.137 N.

Using the Dilution Formula

We can use the dilution formula, which states that the number of equivalents (or moles) before and after dilution remains the same: \(N_1V_1 = N_2V_2\).

• \(N_1\) = Normality of the concentrated source solution (\(\approx 10.137 \text{ N}\)).
• \(V_1\) = Volume of the concentrated source solution needed (this is what we need to find, in mL).
• \(N_2\) = Normality of the target solution (5 N).
• \(V_2\) = Volume of the target solution (1000 mL).

Plugging in the values:

\( 10.137 \text{ N} \times V_1 = 5 \text{ N} \times 1000 \text{ mL} \)

\( V_1 = \frac{5 \times 1000 \text{ mL}}{10.137} \)

\( V_1 = \frac{5000 \text{ mL}}{10.137} \)

\( V_1 \approx 493.24 \text{ mL} \)

The calculated volume is approximately 493.24 mL. This is very close to option 2 (493 mL).

Summary of Calculation

Based on the calculation, approximately 493 mL of the 37% concentrated HCl solution is needed.

The final answer is 493 mL.