Question:
100 mL of 0.01 M HCl is added to 900 mL H\(_2\)O. What is the pH of the resultant solution?
100 mL of 0.01 M HCl is added to 900 mL H\(_2\)O. What is the pH of the resultant solution?
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For strong acids, pH depends only on final concentration of \(H^+\), not on dilution steps separately.
Updated On: Jun 19, 2026
- 1
- 2.7
- 2.3
- 3
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The Correct Option is D
Solution and Explanation
Step 1: Understanding dilution concept.
HCl is a strong acid and completely dissociates in water to give \(H^+\) ions. When it is diluted, the number of moles of \(H^+\) remains constant, but the volume increases, reducing concentration.
Step 2: Calculating moles of HCl initially.
Given: \[ \text{Volume} = 100 \text{ mL} = 0.1 \text{ L}, \quad M = 0.01 \] \[ \text{Moles of HCl} = 0.01 \times 0.1 = 0.001 \text{ mol} \]
Step 3: Finding total volume after dilution.
After adding 900 mL water: \[ V_{total} = 100 + 900 = 1000 \text{ mL} = 1 \text{ L} \]
Step 4: Calculating final concentration of \(H^+\).
Since HCl is strong acid: \[ [H^+] = \frac{0.001}{1} = 10^{-3} \text{ M} \]
Step 5: Calculating pH.
\[ pH = -\log[H^+] = -\log(10^{-3}) = 3 \]
Step 6: Final interpretation.
The solution becomes less acidic after dilution, but still remains acidic with pH equal to 3.
Final Answer: \[ \boxed{3} \]
HCl is a strong acid and completely dissociates in water to give \(H^+\) ions. When it is diluted, the number of moles of \(H^+\) remains constant, but the volume increases, reducing concentration.
Step 2: Calculating moles of HCl initially.
Given: \[ \text{Volume} = 100 \text{ mL} = 0.1 \text{ L}, \quad M = 0.01 \] \[ \text{Moles of HCl} = 0.01 \times 0.1 = 0.001 \text{ mol} \]
Step 3: Finding total volume after dilution.
After adding 900 mL water: \[ V_{total} = 100 + 900 = 1000 \text{ mL} = 1 \text{ L} \]
Step 4: Calculating final concentration of \(H^+\).
Since HCl is strong acid: \[ [H^+] = \frac{0.001}{1} = 10^{-3} \text{ M} \]
Step 5: Calculating pH.
\[ pH = -\log[H^+] = -\log(10^{-3}) = 3 \]
Step 6: Final interpretation.
The solution becomes less acidic after dilution, but still remains acidic with pH equal to 3.
Final Answer: \[ \boxed{3} \]
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