A) \[7\]
B) \[p{{K}_{w}}\]
C) Zero
D) \[1\]
Correct Answer: B
Solution :
At \[{{25}^{o}}C,\]\[{{H}_{2}}O\rightleftharpoons {{H}^{+}}+O{{H}^{-}}\] \[K=\frac{[{{H}^{+}}]\,[O{{H}^{-}}]}{[{{H}_{2}}O]}\] \[K[{{H}_{2}}O]=[{{H}^{+}}][O{{H}^{-}}]\] \[{{K}_{w}}=[{{H}^{+}}]\,[O{{H}^{-}}]\] \[[{{H}^{+}}]\,[O{{H}^{-}}]=1\times {{10}^{-14}}\] On taking log, \[\log [{{H}^{+}}]+\log [O{{H}^{-}}]=1\times {{10}^{-14}}\] \[\because \] \[{{K}_{w}}=1\times {{10}^{-14}}\] or \[\log [{{H}^{+}}]+\log [O{{H}^{-}}]=\log {{K}_{w}}=-14\] \[-\log [{{H}^{+}}]+(-\log [O{{H}^{-}}])=-\log {{K}_{w}}=14\] (at \[{{25}^{o}}C\]) \[pH+pOH=p{{K}_{w}}=14\] \[\because \] \[pH=-\log [{{H}^{+}}]\] \[pOH=-\log [O{{H}^{-}}]\] or \[-\log {{K}_{w}}=p{{K}_{w}}\] \[pH+pOH=p{{K}_{w}}\]
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