A) \[3.8\times {{10}^{-12}}~\]
B) \[3.8\]
C) 6
D) \[13.4\]
Correct Answer: B
Solution :
\[K=\frac{[{{H}_{3}}{{O}^{+}}]\,[HC{{O}_{3}}^{-}]}{[C{{O}_{2}}]\,{{[{{H}_{2}}O]}^{2}}}\]As \[pH=6.0{{[{{H}_{3}}O]}^{+}}={{10}^{-6}}\] \[K=\frac{[{{H}_{3}}{{O}^{+}}]\,[HC{{O}_{3}}^{-}]}{[C{{O}_{2}}]\,{{[{{H}_{2}}O]}^{2}}}\] (\[{{H}_{2}}O\] is in excess, therefore its cone. remains constant) \[\frac{\left[ HC{{O}_{3}}^{-} \right]}{\left[ C{{O}_{2}} \right]}=\frac{K}{\left[ {{H}_{3}}{{O}^{+}} \right]}=\frac{3.8\times {{10}^{-6}}}{{{10}^{-6}}}=3.8\]You need to login to perform this action.
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