A) 0.01 N
B) \[4.45\times {{10}^{-5}}\]
C) \[8.0\times {{10}^{-5}}\]
D) unpredictable
Correct Answer: B
Solution :
\[{{H}_{2}}A{{H}^{+}}+H{{A}^{-}}\] as \[pH=4[{{H}^{+}}]={{10}^{-4}}M\] \[{{K}_{{{a}_{1}}}}=\frac{[{{H}^{+}}][H{{A}^{-}}]}{[{{H}_{2}}A]}\] Or \[\frac{{{10}^{-4}}\times [H{{A}^{-}}]}{{{10}^{-2}}}=4.45\times {{10}^{-7}}\] Or \[[H{{A}^{-}}]=4.45\times {{10}^{-5}}\]You need to login to perform this action.
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