Question:
The pH of \(0.01\;N\) lime water is
The pH of \(0.01\;N\) lime water is
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For strong bases, normality directly gives hydroxide ion concentration. First calculate \(pOH\), then use:
\[
pH=14-pOH
\]
Updated On: Jun 22, 2026
- \(13.09\)
- \(10\)
- \(12\)
- \(9.8\)
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The Correct Option is C
Solution and Explanation
Step 1: Understand what lime water is.
Lime water is an aqueous solution of calcium hydroxide:
\[ Ca(OH)_2 \] It is a strong base and dissociates completely in water as:
\[ Ca(OH)_2 \longrightarrow Ca^{2+} + 2OH^- \]
Step 2: Relate normality with hydroxide ion concentration.
For bases, normality directly gives the concentration of replaceable hydroxide ions.
Given:
\[ 0.01\;N \] Thus, hydroxide ion concentration is:
\[ [OH^-]=0.01=10^{-2} \]
Step 3: Calculate the pOH.
Using the formula:
\[ pOH=-\log[OH^-] \] Substituting the value:
\[ pOH=-\log(10^{-2}) \] \[ pOH=2 \]
Step 4: Calculate the pH.
We know:
\[ pH+pOH=14 \] Therefore:
\[ pH=14-2 \] \[ pH=12 \]
Step 5: Match with the given options.
The correct option is:
\[ (3)\;12 \]
Step 6: Final conclusion.
Hence, the pH of \(0.01\;N\) lime water is:
\[ \boxed{12} \]
Lime water is an aqueous solution of calcium hydroxide:
\[ Ca(OH)_2 \] It is a strong base and dissociates completely in water as:
\[ Ca(OH)_2 \longrightarrow Ca^{2+} + 2OH^- \]
Step 2: Relate normality with hydroxide ion concentration.
For bases, normality directly gives the concentration of replaceable hydroxide ions.
Given:
\[ 0.01\;N \] Thus, hydroxide ion concentration is:
\[ [OH^-]=0.01=10^{-2} \]
Step 3: Calculate the pOH.
Using the formula:
\[ pOH=-\log[OH^-] \] Substituting the value:
\[ pOH=-\log(10^{-2}) \] \[ pOH=2 \]
Step 4: Calculate the pH.
We know:
\[ pH+pOH=14 \] Therefore:
\[ pH=14-2 \] \[ pH=12 \]
Step 5: Match with the given options.
The correct option is:
\[ (3)\;12 \]
Step 6: Final conclusion.
Hence, the pH of \(0.01\;N\) lime water is:
\[ \boxed{12} \]
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