Question:
What is the pH of solution containing \(5\times 10^{-4} \text{M} \text{H}^+\) ion?
What is the pH of solution containing \(5\times 10^{-4} \text{M} \text{H}^+\) ion?
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pH is the negative log of the hydrogen ion concentration; log 5 is 0.699.
Updated On: Oct 1, 2026
- \(4.699\)
- \(3.301\)
- \(5.401\)
- \(3.725\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
pH is defined as \(\text{pH} = -\log[\text{H}^+]\).
Step 2: Calculation:
\[ \text{pH} = -\log(5\times 10^{-4}) = -(\log 5 + \log 10^{-4}) \]
\[ = -(0.699 - 4) = 4 - 0.699 = 3.301 \]
Step 3: Check the Other Options:
4.699 is \(4+0.699\), a sign mistake. 5.401 and 3.725 do not come from any correct handling of the logarithm. As an estimate, \([\text{H}^+] = 5\times10^{-4}\) lies between \(10^{-3}\) (pH 3) and \(10^{-4}\) (pH 4), so the pH must be between 3 and 4, which only 3.301 satisfies. Option (B) is correct.
Final Answer:
The pH is 3.301, option (B). \[ \boxed{\text{(B) } 3.301} \]
pH is defined as \(\text{pH} = -\log[\text{H}^+]\).
Step 2: Calculation:
\[ \text{pH} = -\log(5\times 10^{-4}) = -(\log 5 + \log 10^{-4}) \]
\[ = -(0.699 - 4) = 4 - 0.699 = 3.301 \]
Step 3: Check the Other Options:
4.699 is \(4+0.699\), a sign mistake. 5.401 and 3.725 do not come from any correct handling of the logarithm. As an estimate, \([\text{H}^+] = 5\times10^{-4}\) lies between \(10^{-3}\) (pH 3) and \(10^{-4}\) (pH 4), so the pH must be between 3 and 4, which only 3.301 satisfies. Option (B) is correct.
Final Answer:
The pH is 3.301, option (B). \[ \boxed{\text{(B) } 3.301} \]
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