H3PO4+H2O⇌H3O++H2PO−4 pK1=2.15 H2PO−4+H2O⇌H3O++HPO24− – InterviewSolution

1.

H3PO4+H2O⇌H3O++H2PO−4 pK1=2.15 H2PO−4+H2O⇌H3O++HPO24− pK2=7.20 Hence pH of 0.01 M NaH2PO4 is:

Answer»

$H_{3} P O_{4} + H_{2} O ⇌ H_{3} O^{+} + H_{2} P O_{4}^{-}$ $p K_{1} = 2.15$

$H_{2} P O_{4}^{-} + H_{2} O ⇌ H_{3} O^{+} + H P O_{4}^{2} -$ $p K_{2} = 7.20$

Hence pH of 0.01 M $N a H_{2} P O_{4}$ is:

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