question_answer 1)

Calculate the pH of \[1.0\times {{10}^{-8}}MHCl\]solution.

Answer:

The neutral water has \[[{{H}^{+}}]=1\times {{10}^{-7}}M\] By adding\[1.0\times {{10}^{-8}}M\,HCl\], a concentration of \[1.0\times {{10}^{-8}}\]\[M\,\,{{H}^{+}}\]ions has increased in solution. Thus, total \[[{{H}^{+}}]=(1\times {{10}^{-7}}+1\times {{10}^{-8}})M\] \[=(1\times {{10}^{-7}}+0.1\times {{10}^{-7}})M=1.1\times {{10}^{-7}}M\] \[pH=-\log (1.1\times {{10}^{-7}})=-[log1.1+log{{10}^{-7}}]\] \[=-[0.0414-7.0]=6.9586\]