H _ 3 PO _ 4 + H _ 2 O H _ 3 O ^ + + H _ 2 PO _ 4 ^ - ; pK 1 = 2.15 H _ 2 PO _ 4 ^ - + H…
Chemistry · JEE Advanced · NTA Exams — Equilibrium
\(\mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ; pK1 = 2.15
\(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HPO}_{4}^{2-}\) pK2 = 7.20
Hence pH of 0.01 M NaH2PO4 is :
\(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HPO}_{4}^{2-}\) pK2 = 7.20
Hence pH of 0.01 M NaH2PO4 is :
- 9.35
- 4.675
- 2.675
- 7.350
Answer
(B) 4.675
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