A) 5
B) 7
C) 8
D) 6.98
Correct Answer: D
Solution :
: For\[0.00001\text{ }M\text{ }HCl,\text{ }\!\![\!\!\text{ }{{H}^{+}}]={{10}^{-5}}M\]. When\[HCl\]is diluted thousand fold, \[[{{H}^{+}}]={{10}^{-8}}M\] Dissociation of water, \[{{H}_{2}}O{{H}^{+}}+O{{H}^{-}}\] Conc. if no dissociation C \[{{10}^{-8}}\] 0 Cone. after dissociation\[C-x\] \[({{10}^{-8}}+x)\] \[x\] \[\Rightarrow \] \[{{x}^{2}}+{{10}^{-8}}x-{{10}^{-14}}=0\] \[\Rightarrow \] \[x=9.5\times {{10}^{-8}}\][neglecting the\[-ve\]root] \[x=[O{{H}^{-}}]=9.5\times {{10}^{-8}}mol\text{ }d{{m}^{-3}}\] \[[{{H}^{+}}]=\frac{{{K}_{w}}}{[O{{H}^{-}}]}=\frac{{{10}^{-14}}}{9.5\times {{10}^{-8}}}\] \[[{{H}^{+}}]=1.005\times {{10}^{-7}}mol\,d{{m}^{-3}}\] \[\therefore \]\[pH=-\log [{{H}^{+}}]=6.97\]
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