Question:
The pH of the resultant solution obtained by mixing 20 mL of 0.01 M HCl and 20 mL of 0.005 M \(Ca(OH)_2\) is
The pH of the resultant solution obtained by mixing 20 mL of 0.01 M HCl and 20 mL of 0.005 M \(Ca(OH)_2\) is
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For acid-base mixing, compare equivalents of \(H^+\) and \(OH^-\), not only molarity.
Updated On: Apr 29, 2026
- 2
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- 1
- 7
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The Correct Option is D
Solution and Explanation
Concept:
HCl is a strong acid and \(Ca(OH)_2\) is a strong base. If moles of \(H^+\) and \(OH^-\) are equal, the solution becomes neutral.
Step 1: Calculate moles of \(H^+\).
\[ \text{Moles of HCl}=0.01\times\frac{20}{1000}=2\times10^{-4} \] So, \[ H^+=2\times10^{-4}\ \text{mol} \]
Step 2: Calculate moles of \(OH^-\).
\[ \text{Moles of }Ca(OH)_2=0.005\times\frac{20}{1000}=1\times10^{-4} \] Each \(Ca(OH)_2\) gives 2 \(OH^-\): \[ OH^-=2\times10^{-4}\ \text{mol} \]
Step 3: Conclusion.
Moles of \(H^+\) and \(OH^-\) are equal. Hence solution is neutral: \[ pH=7 \]
Step 1: Calculate moles of \(H^+\).
\[ \text{Moles of HCl}=0.01\times\frac{20}{1000}=2\times10^{-4} \] So, \[ H^+=2\times10^{-4}\ \text{mol} \]
Step 2: Calculate moles of \(OH^-\).
\[ \text{Moles of }Ca(OH)_2=0.005\times\frac{20}{1000}=1\times10^{-4} \] Each \(Ca(OH)_2\) gives 2 \(OH^-\): \[ OH^-=2\times10^{-4}\ \text{mol} \]
Step 3: Conclusion.
Moles of \(H^+\) and \(OH^-\) are equal. Hence solution is neutral: \[ pH=7 \]
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