A) 8
B) 6.9586
C) more than 8
D) slightly more than 7
Correct Answer: B
Solution :
: The neutral water has \[[{{H}^{+}}]=1\times {{10}^{-7}}M\] By add in\[1.0\times {{10}^{-8}}M\text{ }HCl\], a concentration of \[1.0\times {{10}^{-8}}M\,HCl\]ions has increased in solution. Thus, total \[[{{H}^{+}}]=(1\times {{10}^{-7}}+1\times {{10}^{-8}})M\] \[=(1\times {{10}^{-7}}+0.1\times {{10}^{-7}})M\] \[=1.1\times {{10}^{-7}}M\] \[pH=-\log (1.1\times {{10}^{-7}})=-log[1.1+log{{10}^{-7}}]\] \[=-[0.0414-7.0]\] \[=6.9586\]
You need to login to perform this action.
You will be redirected in
3 sec
