A) \[8\]
B) \[-8\]
C) between \[7\] and \[8\]
D) between \[6\] and \[7\]
Correct Answer: D
Solution :
\[{{10}^{-8}}M\,HCl\] \[M=N\]of \[HCl\] So, \[[{{H}^{+}}]=10\times {{10}^{-8}}\] \[pH=-\log [{{H}^{+}}]\] \[pH=-\log {{10}^{-8}}\] \[pH=8\] So \[[{{H}^{+}}]=1\times {{10}^{-8}}\] This is wrong answer because pH value of acid less than 7 so right pH value calculate as follows Total \[[{{H}^{+}}]\] in \[HCl=1\times {{10}^{-8}}m/L\] Total \[[{{H}^{+}}]\] in \[{{H}_{2}}O=1\times {{10}^{-7}}m/L\] Total \[[{{H}^{+}}]=1\times {{10}^{-7}}+0.1\times {{10}^{-7}}=1.1\times {{10}^{-7}}\] \[[{{H}^{+}}]=1\times {{10}^{-7}}+0.1\times {{10}^{-7}}=1.1\times {{10}^{-7}}\] \[[{{H}^{+}}]=11\times {{10}^{-8}}\] \[pH=-\log (11\times {{10}^{-8}})\] \[pH=-\log 11+8\log 10\] \[pH=-1.0414+8\] \[pH=6.958\]
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