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K(a1),K(a2),K(a3) values for H(3)PO(4) a...

`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)))`?

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If K_(a_(1))gtK_(a_(2)) of H_(2)SO_(4) are 10^(-2) and 10^(-6) respectively then:

K_(1) and K_(2)\ of H_(2) S are 10^(-3) and 10^(-12) respectively what is [S^(-2)] in 0.1 M H_(2) S buffered at p(H) =2

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-)) :

The K_(a) values for HPO_(4)^(2-) and HSO_(3)^(-) are 4.8xx10^(-13) and 6.3xx10^(8) repectively. Therefore, it follows the HPO_(4)^(2-) is … acid than HSO_(3)^(-) and PO_(4)^(3-) is a …… base than SO_(3)^(2-)

The pH of blood is 7.4 . What is the ratio of [(HPO_(4)^(2-))/(H_(2)PO_(4)^(-))] in the blood. pK_(a)(H_(2)PO_(4)^(-))=7.1

In a solution of 0.1M H_(3)PO_(4) acid (Given K_(a_(1))=10^(-3),K_(a_(2))=10^(-7),K_(a_(3))=10^(-12)) Concentration of H_(2)PO_(4)^(-) is

For H_(3)PO_(4) , H_(3)PO_(4) rarr H_(2)PO_(4)^(-)+H^(+)(K_(1)), H_(2)PO_(4)^- rarr HPO_(4)^(-)+H^(+) (K_2), HPO_(4)^(2-) rarr PO_(4)^(3-) + H^(+) (K_(3)) then

If K_(a_1),K_(a_2) and K_a_3) be the first, second and third dissociation constant of H_(3)PO_(4) and K_(a_1)gtgt K_(2_a) gtgtK_(a_3) whis is/are correct :

H_(4)underline(P_(2))O_(6)+H_(2)O to H_(3)PO_(3)+H_(3)PO_(4)

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