1.

A weak acid HA (50.0ml) was titrated against 0.1 M NaOH. The pH values when 20 ml & 40 ml base have been added are found to be 4.898 & 5.324 respectively. Calculate the pH of the solution at equivalence point. (Given : log8//3=0.426,log2=0.301,log15=1.176) Mark your answer in single digit to the nearest integral digit, say your answer is 7.213 then mark 0007 or if your answer is 12.567 then mark as 0013.

Answer»


Solution :`4.898=pk_(a)+"log"(2)/(a-2)`
`5.324=pk_(a)+"log"(4)/(a-4)`
`0.426=log[((4)/(a-4))xx((a-2)/(2))]`
`2.66686=(2(a-2))/((a-4))`
`(8)/(3)=(2(a-2))/(a-4)`
`4(a-4)=3(a-2)`
`a=10`
`4.898=pk_(a)+"log"(2)/(8)`
`pk_(a)=505`
`pH=7+(1)/(2)(5.5+"log"(10)/(150))`
`=9.162` ANS.


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