Q.7 If 50 mL of 0.02 M HCl is added to 950 mL of H2O, then the pH of the final solution will
be________
Diluting 50 mL of 0.02 M HCl with 950 mL of water yields a final
pH of 3.00. Since HCl is a strong acid, it fully dissociates in water, and the pH
can be calculated directly from the hydrogen ion concentration after dilution.
Step-by-Step Calculation
HCl acts as a strong acid and ionizes completely:
HCl → H+ + Cl−
The key principle used here is conservation of moles during dilution.
1. Initial moles of H+
0.02 M × 0.050 L = 0.001 mol H+
2. Total volume after dilution
50 mL + 950 mL = 1000 mL = 1.0 L
3. Final hydrogen ion concentration
[H+] = 0.001 mol / 1 L = 0.001 M = 10−3 M
4. pH calculation
pH = −log(10−3) = 3.00
The autoionization of water (10−7 M) is ignored here because it is negligible compared
to 0.001 M.
Why Options Analysis (Multiple-Choice Style)
| Option | Description | Resulting pH | Correct? | Reason |
|---|---|---|---|---|
| A | No dilution considered | 1.70 | No | Uses original 0.02 M concentration without accounting for volume increase |
| B | Incorrect mole calculation | 2.00 | No | Assumes 0.01 mol instead of the correct 0.001 mol |
| C | Correct dilution calculation | 3.00 | Yes | Correct moles and final volume used |
| D | Over-dilution error | 3.30 | No | Uses incorrect concentration after dilution |
| E | Assumes neutrality | 7.00 | No | Ignores the contribution of HCl entirely |
Applications in Chemistry
This dilution concept is widely used in titrations, buffer preparation, analytical chemistry,
and industrial pH control. For extremely dilute acids (below 10−6 M), water’s
autoionization must be considered using a quadratic equation. However, that correction is unnecessary
in this problem.
