Question 54

An analyst wants to convert $$1$$ L HCl of pH $$= 1$$ to a solution of HCl of pH $$= 2$$. The volume of water needed to do this dilution is _____ mL. (Nearest integer)

Correct Answer: 900

Solution

Detailed Solution

Problem Statement: An analyst wants to convert 1 L HCl of pH = 1 to a solution of HCl of pH = 2. The volume of water needed to do this dilution is ________ mL. (Nearest integer)

Core Theoretical Principles

Hydrochloric acid (HCl) is a strong monobasic acid that dissociates completely in an aqueous solution. Therefore, the molar concentration of hydronium ions [H+] is directly equal to the total molarity of the HCl solution (M).

The relationship between pH and hydrogen ion concentration is defined by the core logarithmic function:

pH = -log10[H+] µ [H+] = 10-pH

When a solution is diluted by adding a pure solvent (water), the total number of moles of solute remains invariant. This conservation of solute quantity is governed by the classical dilution law equation:

M 1 V 1 = M 2 V 2

Where M1 and V1 represent the initial molar concentration and volume, and M2 and V2 represent the final molar concentration and volume post-dilution.

Step-by-Step Derivation

Step 1: Extract Initial Parameters
For the initial state of the hydrochloric acid solution:

Initial Volume (V1) = 1 L = 1000 mL
Initial pH = 1
Initial Concentration (M1) = 10−1 M = 0.1 M

Step 2: Extract Target Parameters
For the final state following the dilution process:

Target pH = 2
Target Concentration (M2) = 10−2 M = 0.01 M

Step 3: Calculate Total Final Volume (V2)
Applying the values to the dilution principle formula:

(0.1 M) \times (1000 mL) = (0.01 M) \times V 2

Rearranging the parameters algebraically to isolate and solve for V2:

V 2 = 0.1 \times 1000 0.01 = 10,000 mL

The total final volume of the solution must become 10,000 mL (which is equivalent to 10 L).

Step 4: Determine Added Solvent Volume (ΔV)
The net volume of extra water needed to expand the system is the absolute difference between the total final volume and the initial volume:

V water = V 2 - V 1

V water = 10,000 mL - 1,000 mL = 9,000 mL

Final Analytical Conclusion

To reduce the concentration exactly tenfold (shifting the pH scale cleanly from 1 to 2), the system volume must scale up by a factor of 10. Starting with an initial volume of 1000 mL, an additional 9000 mL of pure water must be mixed into the solution.

Correct Numerical Value: 9000

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