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-))`:

Options

A. `4.7,9.6`
B. `9.6,4.7`
C. `9.3,4.4`
D. `4.4,9.3`

Explanation

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`