A) 6.23
B) 9.22
C) 7.21
D) 8.95
Correct Answer: D
Solution :
\[\underset{2}{\mathop{NaOH}}\,+\underset{2}{\mathop{HA}}\,\to \underset{-}{\mathop{NaA}}\,+\underset{-}{\mathop{{{H}_{2}}O}}\,\] At end point \[\equiv \] 0.1 x 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. \[C=[NaA]=\frac{2}{20+20}=0.05M\] 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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