A) \[HA:{{K}_{a}}=1\times {{10}^{-8}}\]
B) \[HB:{{K}_{a}}=2\times {{10}^{-6}}\]
C) \[HC:{{K}_{a}}=3\times {{10}^{-8}}\]
D) \[HD:{{K}_{a}}=4\times {{10}^{-10}}\]
Correct Answer: D
Solution :
\[{{K}_{h}}=\frac{{{K}_{w}}}{{{K}_{a}}}\] smaller the \[{{K}_{a}},\]larger the \[{{K}_{h}}\] and thus, higher the hydrolysis.
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