A) 5.0
B) 6.0
C) 7.0
D) 9.0
Correct Answer: C
Solution :
pKa = ? log Ku, pKb = ? log Kb pH = \[-\frac{1}{2}[\log {{K}_{a}}+\log {{K}_{w}}-\log {{K}_{b}}]\] \[=-\frac{1}{2}[-5+\log (1\times {{10}^{-14}})-(-5)]\] \[=-\frac{1}{2}[-5-14+5]=-\frac{1}{2}(-14)=7\]You need to login to perform this action.
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