A) \[7.04\]
B) \[7.40\]
C) \[6.89\]
D) \[6.0\]
Correct Answer: A
Solution :
| The neutral water has \[[O{{H}^{-}}]=1\times {{10}^{-7}}M\] |
| By adding \[{{10}^{-8}}M\,\,NaOH,\] a concentration of \[{{10}^{-8}}M\,\,O{{H}^{-}}\] has increased in solution. |
| Thus, total \[[O{{H}^{-}}]={{10}^{-8}}+{{10}^{-7}}={{10}^{-8}}+10\times {{10}^{-8}}\]\[=11\times {{10}^{-8}}\] |
| \[pOH=-\log 11\times {{10}^{-8}}=6.9586\] |
| \[pH=14-6.9586=7.04\] |
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