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pH of 0.01 M `(NH_(4))_(2) SO_(4)` and 0.02 M `NH_(4)OH` buffer `(pK_(a)` of `NH_(4)^(+)=9.26)` isA. `4.74+log 2 `B. `4.74 -log2`C. `9.26+log 2`D. `9.26+log 1` |
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Answer» Correct Answer - D `pK_(a)(NH_(4)^(+))=9.26` `pK_(b)(NH_(3))=14-9.26=4.74` `[NH_(4)]=2xx[(NH_(4))_(2)SO_(4)]` `=2xx0.01=0.02 M` `pOH =pK_(b)+log . ([NH_(4)^(+)])/([NH_(4)OH])` `pOH =4.74 +log . (0.02)/(0.02)=4.74-log 1` `pH =14-pOH =1404.74+log 1` `=9.26+log 1` |
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