A) 4
B) 10
C) 7
D) 14
Correct Answer: A
Solution :
Given, \[{{K}_{b}}={{10}^{-10}}\] According to question \[{{X}^{-}}+{{H}_{2}}OO{{H}^{-}}+HX\] ... (i) \[H\,X{{H}^{+}}+{{X}^{-}}\] ... (ii) \[\therefore \] \[{{K}_{b}}=\frac{[O{{H}^{-}}]\,[HX]}{[{{X}^{-}}]}\,\,(\because \,[{{H}_{2}}O]=1)\] \[{{K}_{a}}=\frac{[{{H}^{+}}]\,[X]}{[HX]}\] ... (iv) Multiply equation (iii) and equation (iv) \[{{K}_{a}}\times {{K}_{b}}=[{{H}^{+}}]\,[O{{H}^{-}}]\] or \[{{K}_{a}}\times {{10}^{-10}}={{10}^{-14}}\] \[\therefore \] \[{{K}_{a}}={{10}^{-4}}\] \[\therefore \] \[pH=pKa=4\] (\[\because \][salt} = [acid]
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