A) neutral
B) basic
C) unaffected
D) more acidic
Correct Answer: A
Solution :
\[pH=5\] \[[{{H}_{3}}{{O}^{+}}]={{10}^{-pH}}={{10}^{-5}}M\] \[{{M}_{1}}={{10}^{-5}}\]let \[{{V}_{1}}=x\] \[{{M}_{2}}=?\,{{V}_{2}}=100x\] According to dilution law \[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\] \[{{10}^{-5}}x={{M}_{2}}=100x\] \[\frac{{{10}^{-5}}x}{100x}={{M}_{2}}\] \[={{10}^{-7}}M={{M}_{2}}\] \[[{{H}_{3}}{{O}^{+}}]={{10}^{-7}}M+{{10}^{-7}}M\] (from water) \[=2\times {{10}^{-7}}M\] \[pH=-\log [{{H}_{3}}{{O}^{+}}]\] \[=-\log 2\times {{10}^{-7}}=6.69\] Hence, the solution becomes almost neutral
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