A) \[8.0\]
B) \[7.0\]
C) \[6.98\]
D) \[14.0\]
Correct Answer: C
Solution :
\[{{10}^{-8}}M\] \[HCl\] Molarity of \[HCl\] = normality of \[HCl\] \[N=1\times {{10}^{-8}}\] or \[[{{H}^{+}}]=1\times {{10}^{-8}}mol/L\] \[pH=-\log [{{H}^{+}}]\] \[pH=-\log {{10}^{-8}}\] \[pH=8\] This is wrong answer because pH of acids less than 7. So, right pH is calculated as follows Total \[[{{H}^{+}}]\] in \[HCl=1\times {{10}^{-8}}mol/L\] Total \[[{{H}^{+}}]\] in \[{{H}_{2}}O=1\times {{10}^{-7}}\] Total \[[{{H}^{+}}]=1\times {{10}^{-7}}+0.1\times {{10}^{-7}}\] \[=1.1\times {{10}^{-7}}\] \[[{{H}^{+}}]=11\times {{10}^{-8}}\] \[\Rightarrow \] \[pH=-\log [{{H}^{+}}]\] \[pH=-\log (11\times {{10}^{-6}})\] \[=-\log 11-\log {{10}^{-8}}\] \[=-1.0414+8\times 1\] \[\therefore \] \[pH=6.9586\]
You need to login to perform this action.
You will be redirected in
3 sec
