A) 6.23
B) 9.22
C) 7.21
D) 8.95
Correct Answer: D
Solution :
\[\underset{2}{\mathop{NaOH}}\,+\underset{2}{\mathop{HA}}\,\xrightarrow{{}}\underset{-}{\mathop{NaA}}\,+H\underset{-}{\mathop{_{2}}}\,O\] At end point \[=0.1\times 20=2\] \[\because \,\,20\,\,mL\] of \[NaOH\] is required for the complete neutralisation of HA. NaA is a salt of strong base and weak acid. Thus, will undergo hydrolysis and solution will become basic. \[C=[NaA]=\frac{2}{20+20}=0.05\,M\] and \[p{{K}_{a}}=-\log \,(6\times {{10}^{-6}})=5.2\] pH at the end point \[=7+\frac{1}{2}\,(p{{K}_{a}}+\log \,C)\] \[7+\frac{1}{2}(5.2+\log \,0.05)=8.95\]You need to login to perform this action.
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