Question:
How many times a 0.1 M strong monobasic acid solution should be diluted so that pH of the resulting solution is tripled?
How many times a 0.1 M strong monobasic acid solution should be diluted so that pH of the resulting solution is tripled?
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Each increase of 1 in pH ⇒ concentration decreases by factor 10.
Updated On: May 2, 2026
- 50
- 10
- 25
- 100
- 1000
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The Correct Option is D
Solution and Explanation
Concept: Relation between pH and concentration
\[ \text{pH} = -\log [H^+] \] For strong acid: \[ [H^+] = \text{concentration} \] ---
Step 1: Initial pH
\[ [H^+] = 0.1 = 10^{-1} \] \[ \text{pH} = 1 \] ---
Step 2: Final pH
Tripled: \[ \text{pH} = 3 \] \[ [H^+] = 10^{-3} \] ---
Step 3: Dilution factor
\[ \text{Dilution factor} = \frac{10^{-1}}{10^{-3}} = 10^2 = 100 \] Actually: \[ \text{Initial pH} = 1 \] \[ \text{Final pH} = 3 \] \[ \Delta pH = 2 \Rightarrow \text{concentration reduced by } 10^2 = 100 \] But "tripled pH" means: \[ 1 \rightarrow 3 \Rightarrow \text{factor} = 100 \] Final Answer: \[ \boxed{100} \]
\[ \text{pH} = -\log [H^+] \] For strong acid: \[ [H^+] = \text{concentration} \] ---
Step 1: Initial pH
\[ [H^+] = 0.1 = 10^{-1} \] \[ \text{pH} = 1 \] ---
Step 2: Final pH
Tripled: \[ \text{pH} = 3 \] \[ [H^+] = 10^{-3} \] ---
Step 3: Dilution factor
\[ \text{Dilution factor} = \frac{10^{-1}}{10^{-3}} = 10^2 = 100 \] Actually: \[ \text{Initial pH} = 1 \] \[ \text{Final pH} = 3 \] \[ \Delta pH = 2 \Rightarrow \text{concentration reduced by } 10^2 = 100 \] But "tripled pH" means: \[ 1 \rightarrow 3 \Rightarrow \text{factor} = 100 \] Final Answer: \[ \boxed{100} \]
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