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pH of `0.1M Na_(2)HPO_(4)` and `0.2M NaH_(2)PO_(4)` are respectively: `(pK_(a)"for" H_(3)PO_(4)` are `2.12, 7.21` and `12.0` for respective dissociation to `HPO_(4)^(2-), HPO_(4)^(-)` and `PO_(4)^(3-))`:A. `4.7,9.6`B. `9.6,4.7`C. `9.3,4.4`D. `4.4,9.3` |
Answer» For `Na_(2)HPO_(4),pH=(pK_(a_(2))+pK_(a_(3)))/(2)=(7.28+12)/(2)=9.6` For `Na_(2)HPO_(4),pH=(pK_(a_(1))+pK_(a_(2)))/(2)=(2.2+7.2)/(2)=4.7` |
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