Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The pH of the resultant solution of 20 mL of 0.1 M H3PO4 and 20 mL of 0.1 M Na3PO4 is:Option: 1 <p

The pH of the resultant solution of 20 mL of 0.1 M H3PO4 and 20 mL of 0.1 M Na3PO4 is:

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.

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Mains 2026 Exam City Slip LIVE: NTA session 2 test centre allotment soon; admit card date
anam-khan

JEE Main 2026: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA session 2 city slip soon; admit card, exam dates, here's how to prepare

Latest Question