A) 10, 6, 2
B) 11, 3, 7
C) 10, 3, 7
D) 10, 4, 7
Correct Answer: B
Solution :
Key Idea: Use \[pH=-\log [{{H}^{+}}]\] and\[pH-\text{ }pOH=14\]to solve problem. pH of \[{{10}^{-3}}M\,NaOH\] \[\underset{{{10}^{-3}}M}{\mathop{NaOH}}\,\xrightarrow[{}]{{}}N{{a}^{+}}+\underset{{{10}^{-3}}M}{\mathop{O{{H}^{-}}}}\,\] \[pOH=-\log {{10}^{-3}}\] \[=3\] \[pH+pOH=14\] \[pH=14-3=11\] \[\mathbf{PH}\]of\[\mathbf{1}{{\mathbf{0}}^{\mathbf{-3}}}\mathbf{HCl}\] \[\underset{{{10}^{-3}}M}{\mathop{HCl}}\,\xrightarrow[{}]{{}}\underset{{{10}^{-3}}M}{\mathop{{{H}^{+}}}}\,+C{{l}^{-}}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log [{{10}^{-3}}]\] \[=3\] pH of\[{{10}^{-3}}NaCl\]is 7 (neutral) \[\because \]It is salt of strong acid and strong base.
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