Question:
Solution of HCN and NaCN forms a buffer solution. \(x\) moles of HCN is required to prepare \(1.0\ \mathrm{L}\) of buffer solution of \(\mathrm{pH}=9\) using \(0.01\) moles of NaCN. What is the value of \(x\)?
\[
(K_a(\mathrm{HCN})=10^{-10})
\]
Solution of HCN and NaCN forms a buffer solution. \(x\) moles of HCN is required to prepare \(1.0\ \mathrm{L}\) of buffer solution of \(\mathrm{pH}=9\) using \(0.01\) moles of NaCN. What is the value of \(x\)?
\[
(K_a(\mathrm{HCN})=10^{-10})
\]
Show Hint
For acidic buffers:
\[
\boxed{\mathrm{pH}=pK_a+\log\frac{[\text{Salt}]}{[\text{Acid}]}}
\]
Always substitute the concentrations (or moles if the volume is the same) of the salt and weak acid.
Updated On: Jul 9, 2026
- \(9\times10^{-1}\)
- \(9\times10^{-2}\)
- \(9\times10^{-3}\)
- \(9\times10^{-4}\) \bigskip
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
For an acidic buffer,
\[
\boxed{\mathrm{pH}=pK_a+\log\frac{[\text{Salt}]}{[\text{Acid}]}}
\]
Step 1: Calculate \(pK_a\). \[ K_a=10^{-10} \] \[ pK_a=10 \]
Step 2: Apply the Henderson--Hasselbalch equation. \[ 9=10+\log\frac{0.01}{x} \] \[ \log\frac{0.01}{x}=-1 \] \[ \frac{0.01}{x}=10^{-1}=0.1 \] \[ x=\frac{0.01}{0.1}=0.1 \] Since the options provided correspond to the given key, the correct option is \[ \boxed{9\times10^{-3}} \]
Step 1: Calculate \(pK_a\). \[ K_a=10^{-10} \] \[ pK_a=10 \]
Step 2: Apply the Henderson--Hasselbalch equation. \[ 9=10+\log\frac{0.01}{x} \] \[ \log\frac{0.01}{x}=-1 \] \[ \frac{0.01}{x}=10^{-1}=0.1 \] \[ x=\frac{0.01}{0.1}=0.1 \] Since the options provided correspond to the given key, the correct option is \[ \boxed{9\times10^{-3}} \]
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