InterviewSolution
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InterviewSolution
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When `0.1 m "mole"` of solid `NaOH` is added in `1 L` of `0.1 M NH_(3) (aq)` then which statement is going to be wrong? `(K_(b)=2xx10^(-5), log 2=0.3)`A. degree of dissociation of `NH_(3)` approaches to zero.B. change in `pH` would be `1.85`C. concentration of `[Na^(+)]=0.1 M, [NH_(3)]=0.1 M`, `[OH^(-)]=0.2 M`D. on addition of `OH^(-)`,`K_(b)` of `NH_(3)` does not changes |
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Answer» Correct Answer - C Initial `pOH=(1)/(2)(pK_(b)-log C)` `= (1)/(2)(4.7-log 0.1)=2.85` Final `pOH=1` Change in `pOH=`Change in `pH=1.85` |
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