Calculate the concentration of NH3andNH+4Cl− present in the buffer solution of pH = 9, when total concentration of buffering reagents is 0.6 molL−1. Take pKb for NH3=4.7,log2=0.3
A
[NH+4Cl−]=0.1M and[NH3]=0.3M
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B
[NH+4Cl−]=0.6M and[NH3]=0.6M
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C
[NH+4Cl−]=0.8M and[NH3]=0.2M
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D
[NH+4Cl−]=0.4M and[NH3]=0.2M
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Solution
The correct option is D[NH+4Cl−]=0.4M and[NH3]=0.2M Given, pKbofNH3=4.7,pH=9 total concentration of buffer = 0.6 molL−1. Let [NH3]=b,[NH+4Cl−]=a ∴[NH3]+[NH+4Cl−]=a+b Using the Henderson-Hasselbalch equation: pOH=pKb+log[[conjugate acid][base]] pOH=pKb+log[NH+4Cl−][NH3]⇒14−pH=4.7+logab ⇒14−pH=4.7+logab ⇒14−9=4.7+logab ⇒14−9=4.7+logab ⇒5=4.7+logab ⇒logab=0.3 ⇒ab=2∴a=2b Given a+b=0.6;2b+b=0.6 ∴3b=0.6orb=0.2mol and a+b=0.6mol a+0.2=0.6⇒a=0.4mol Thus, [NH+4Cl−]=0.4M and[NH3]=0.2M