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A buffer solution 0.04 M in Na(2)HPO(4) ...

A buffer solution 0.04 M in `Na_(2)HPO_(4)` and 0.02 in `Na_(3)PO_(4)` is prepared. The electrolytic oxidation of 1.0 milli-mole of the organic compound RNHOH is carried out in 100 mL of the buffer. The reaction is `RNHOH+H_(2)OrarrRNO_(2)+4H^(+)+4e^(-)` The approximate pH of solution after the oxidation is complete is :
`[Given : for H_(3)PO_(4),pK_(a1)=2.2,pK_(a2)=7.20,pK_(a3)=12]`

A

`6.90`

B

`7.20`

C

`7.5`

D

None of these

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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.

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 pH of the resultant solution of 20 mL of 0.1 M H_(3)PO_(4) and 20 mL of 0.1 M Na_(3)PO_(4) is :

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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)

What will be the pH of a solution formed by mixing 10 ml 0.1 M NaH_(2)PO_(4) and 15 mL 0.1 M Na_(2)HPO_(4) ? ["Given: for "H_(3)PO_(4)Pk_(a_(1))=2.12, Pk_(a_(2))=7.2]

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NARENDRA AWASTHI-IONIC EEQUILIBRIUM-Exercise
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