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H(3)PO(4)+H(2) Leftrightarrow H(3)O^(+)+...

`H_(3)PO_(4)+H_(2) Leftrightarrow H_(3)O^(+)+H_(2)PO_(4)^(-),pK_(1)=2.15`
`H_3PO_(4)^(-)+H_(2)O Leftrightarrow H_(3)O^(+)+HPO_(4)^(2-), pK_(2)=7.20`
Hence pH of 0.01 M `NaH_(2)PO_(4)` is

A

9.35

B

4.675

C

2.675

D

`7.350`

Text Solution

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The correct Answer is:
B
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