H3PO4+H2O⇌H3O++H2PO−4; pK1=2.15 H2PO−4+H2O⇌H3O++HPO−24; – InterviewSolution

1.	H3PO4+H2O⇌H3O++H2PO−4; pK1=2.15 H2PO−4+H2O⇌H3O++HPO−24; pK2=7.20 Hence, pH of 0.01 M NaH2PO4 is:
Answer» $H_{3} P O_{4} + H_{2} O ⇌ H_{3} O^{+} + H_{2} P O_{4}^{-}; p K_{1} = 2.15$ $H_{2} P O_{4}^{-} + H_{2} O ⇌ H_{3} O^{+} + H P O_{4}^{- 2}; p K_{2} = 7.20$ $H e n c e, p H o f 0.01 M N a H_{2} P O_{4} i s$ :

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