What is the equilibrium constant for the reaction HB(aq.)+NaA(aq.)⇌HA(aq.)+NaB(aq.)?
A
10
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B
0.1
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C
10−7
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D
10−11
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Solution
The correct option is D10 As we know, at half point of equivalence point pH=pKa and from figure it can be seen that at half point pH=5 for HB so its pKa is 5. Similarly pKa is 6 for HA. For this reaction, K=KHB/KHA=10−5/10−6=10