A) 13.30
B) 14.70
C) 12.30
D) 12.95
Correct Answer: C
Solution :
For \[N{{H}_{4}}OH\]. \[[O{{H}^{-}}]=C\,.\,\alpha \] ; \[C=\frac{1}{10}M\], \[\alpha =0.2\] \[[O{{H}^{-}}]=\frac{1}{10}\times 0.2=2\times {{10}^{-2}}\,M\] \[pOH=-\log \,\,[O{{H}^{-}}]\]\[=\log \,\,[2\times {{10}^{-2}}]\]; \[pOH=1.7\] \[pH=14-pOH\] \[=14-1.7=12.30\].You need to login to perform this action.
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