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The pH of the resultant solution of 20 mL of 0.1 M H3POand 20 mL of 0.1 M Na3POis:

Option: 1

pK_{a_{1}}


Option: 2

pK_{a_{2}}
 


Option: 3

\mathrm{\frac{pK{a_{{_{^{_{1}}}}}}+pK_{a_{2}}}{2}}


Option: 4

2


Answers (1)

best_answer

\\\mathrm{H_{3}PO_{4}\: +\: PO_{4}^{3-}\: \rightarrow \: H_{2}PO_{4}^{-}\: +\: HPO_{4}^{2-}}\\\\\mathrm{[H_{2}PO_{4}^{-}]\: =\: [HPO_{4}^{2-}]}\\\\\mathrm{The\: resultant\: solution\: will\: be\: a\: buffer\: solution.}\\\\\mathrm{\therefore pH\: =\: pK_{a_{2}}\: +\: log\left [\frac{HPO^{2-}_{4}}{H_{2}PO_{4}^{-}}\right ] }\\\mathrm{from\: above,\: pH\:=\: pK_{a_{2}}}

Therefore option (2) is correct.

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avinash.dongre

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