A) 10
B) 7
C) 6
D) 4
Correct Answer: D
Solution :
Key Idea (i) For basic buffer \[pOH=p{{K}_{b}}+\log \frac{[salt]}{[base]}\] (ii)\[pH+pOH=14\] Given, \[{{K}_{b}}=1\times {{10}^{-10}},[salt]=[base]\] \[pOH=-\log \,{{K}_{b}}+\log \frac{[salt]}{[base]}\] \[\therefore \] \[pOH=-\log (1\times {{10}^{-10}})+\log 1\] \[=10\] \[pH+pOH=14\] \[pH=14-10=4\]You need to login to perform this action.
You will be redirected in
3 sec