A) \[3:17\]
B) \[10:0\]
C) \[3:0\]
D) \[6:0\]
Correct Answer: B
Solution :
[b] Buffer A \[p{{H}_{A}}=p{{K}_{a}}+\log \left( \frac{a}{b} \right)\] ?..(i) Buffer B \[p{{H}_{B}}=p{{K}_{a}}+\log \left( \frac{b}{a} \right)\] ??(ii) Since \[a>b\] \[\therefore \,\,\,\,p{{H}_{A}}-p{{H}_{B}}=\log \frac{a}{b}-\log \frac{b}{a}\] \[2=\log \frac{a}{b}+\log \frac{a}{b}\] \[\Rightarrow \,\,\,\,\,2=2\log \frac{a}{b}\] \[\therefore \,\,\,\,\,\log \frac{a}{b}=1\] \[\frac{a}{b}=\]Antilog \[(1)=10\] \[\therefore \,\,\,\,\,\frac{a}{b}=10\]
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