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-))`:

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 H_2PO_(4)^(-), HPO_(4)^(2-) and PO_(4)^(3-)) :

pH of 0.1 M Na_2HPO_4 and 0.2 M NaH_2PO_4 are respectively : ( pK_a for H_3PO_4 are 2.2,7.2,12.0)

Calculate pH a) NaH_(2)PO_(4) b) Na_(2)HPO_(4) respectively, for H_(3)PO_(4) pKa_(1) = 2.25, pKa_(2) = 7.20, pKa_(3) = 12.37 )

The addition of NaH_(2)PO_(4) to 0.1M H_(3)PO_(4) will cuase

The addition of NaH_(2)PO_(4) to 0.1M H_(3)PO_(4) will cuase

K_(a1),K_(a2),K_(a3) values for H_(3)PO_(4) are 10^(-3),10^(-8) and 10^(-12) respectively .If K_(w)(H_(2)O)=10^(-14) then , (i) What is dissociation constant of HPO_(4)^(2-) ? (ii) What is K_(b) of HPO_(4)^(2-) (iii) What is K_(b) of H_(2)PO_(4)^(-) ? (iv) What is order of K_(b) of Po_(4)^(3-)(K_(b_(3))),HPO_(4)^(2-)(K_(b_(2))) and H_(2)PO_(4)^(-)(K_(b_(1))) ?

The two Bronsted bases in the reaction are HC_(2)O_(4)^(-)+PO_(4)^(3-) to HPO_(4)^(2-)+C_(2)O_(4)^(2-)

A buffer solution of 0.080M Na_(2)HPO_(4) and 0.020 M Na_(3)PO_(4) is prepared. The electrolytic oxidation of 1.0 mmol RNHOH is carried out in 100mL buffer to give RNHOH + H_(2)O rarr RNO_(2) + 4H^(o+) + 4e^(-) Calculate approximate pH of the solution after oxidation is complete pK_(a_(2)), pK_(a_(2)) , and pK_(a_(3)) of H_(3)PO_(4) are 2.12,7.20 , and 12.0 , respectively.

The anhydride of acid H_(3)PO_(4) and HPO_(3) are: