A) 12.98
B) 12.13
C) 10.48
D) 9.24
Correct Answer: B
Solution :
| [b] After electrolysis aqueous NaCl is converted into aqueous NaOH. The quantity of electricity passed |
| \[=\frac{0.250\times 35\times 60}{96500C}A\,s=5.44\times {{10}^{-3}}F\] |
| The number of equivalents of \[O{{H}^{-}}\] ion formed |
| \[=5.44\times {{10}^{-3}}\] |
| \[\therefore \] Molarity of \[NaOH=\frac{5.44\times {{10}^{-3}}}{0.400L}eq.\] |
| \[=1.36\times {{10}^{-2}}\] |
| \[pOH=-\log (1.36\times {{10}^{-2}})=2.00-0.13=1.87\] |
| \[\therefore \] \[pH=12.13\] |
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