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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. |
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Answer» `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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