A) 4
B) 9.0
C) 10
D) 9.6
Correct Answer: D
Solution :
[d] Given at 330 K \[{{K}_{w}}={{10}^{-13.6}}\] i.e. \[p{{K}_{w}}=pH+pOH\] \[\therefore \text{ }pOH=-log\left[ O{{H}^{-}} \right]\] \[13.6=pH+pOH\] \[pOH=-log{{10}^{-4}}\] \[pOH=4\] \[\therefore pH=13.6-4=9.6\]You need to login to perform this action.
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