A) \[6\]
B) \[7\]
C) \[8\]
D) \[6.98\]
Correct Answer: D
Solution :
\[HCl(aq)+{{H}_{2}}O(l){{H}_{3}}{{O}^{+}}(aq)+C{{l}^{-}}(aq)\] \[2{{H}_{2}}O(l){{H}_{3}}{{O}^{+}}(aq)+O{{H}^{-}}(aq)\] Sources of \[{{H}_{3}}{{O}^{+}}\] ions are both \[HCl\] and \[{{H}_{2}}O\]. Hence, \[[{{H}_{3}}{{O}^{+}}]={{10}^{-6}}+x\] \[\because \] \[{{K}_{w}}={{10}^{-14}}\] \[\therefore \] \[{{10}^{-14}}=[{{10}^{-8}}+x][x]\] or \[{{x}^{2}}+{{10}^{-8}}x-{{10}^{-14}}=0\] or \[x=9.5\times {{10}^{-8}}\] or \[x=[O{{H}^{-}}]9.5\times {{10}^{-8}}\] \[\therefore \] \[pOH=7.02\]and \[pH=6.98\]
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