A) 4
B) -10
C) 4
D) 7
Correct Answer: D
Solution :
Given, concentration of\[NaOH={{10}^{-10}}M\] \[\underset{{{10}^{-10}}}{\mathop{NaOH}}\,\xrightarrow{{}}\underset{{{10}^{-10}}}{\mathop{N{{a}^{+}}}}\,+\underset{{{10}^{-10}}}{\mathop{O{{H}^{-}}}}\,\] \[\therefore \] \[{{[OH]}^{-}}\]from\[NaOH={{10}^{-10}}\] \[\therefore \] Total\[[O{{H}^{-}}]={{10}^{-7}}+{{10}^{-10}}\] \[={{10}^{-7}}(1+0.001)\] \[={{10}^{-7}}\left( \frac{1001}{1000} \right)\] \[={{10}^{-10}}\times 1001\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log (1001\times {{10}^{-10}})\] \[=-3.004+10\] \[=6.9996\] \[pH+pOH=14\] \[\therefore \] \[pH=14-6.9996\] \[=7.0004\approx 7\]
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