A) 11
B) 8
C) 6
D) 14
Correct Answer: A
Solution :
[a] \[C{{O}_{2}}\] with \[{{H}_{2}}O\] forms \[{{H}_{2}}C{{O}_{3}}\] \[C{{O}_{2}}+{{H}_{2}}O\rightleftharpoons {{H}^{+}}+HCO_{3}^{-}\] \[{{K}_{1}}=\frac{[{{H}^{+}}][HCO_{3}^{-}]}{[C{{O}_{2}}]}=4.5\times {{10}^{-7}}\] Again \[pH=-\log [{{H}^{+}}]=7.4\] \[\therefore \,[{{H}^{+}}]=4.0\times {{10}^{-8}}\] \[\therefore \,\frac{[HCO_{3}^{-}]}{[C{{O}_{2}}]}=\frac{4.5\times {{10}^{-7}}}{4\times {{10}^{-8}}}=11\]You need to login to perform this action.
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