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(i) Derive Henderson- Hasselbalchequation (ii) Whatis kohlraush's law ? |
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Answer» Solution :(i) The concentration of hydronium ion in an acidic buffersolution dependson the ratioof the concentrationof the weakacidto the concentration of itsconjugatebasepresentin the solutioni.e, ` [ H_(3)O^(+)] = K_(a)([" ACID"]_(aq))/(["base"]_(aq))` (i) The weakacidis dissociatedonlyto small extent . Moreover dueto commonion effect,the dissociationis further SUPPRESSED and hencethe equilibriumconcentrationof the acidis nearly equal to the initialconcentration of the unionisedacid.Similarly the concentrationof the conjugatebaseis nearlyequalto the initial concentration of the addedsalt. ` [H_(3)O^(+) K_(a)] ("[ acid"])/("[salt"])` (ii) Here[ acid ] and [salt] represent the initial concentration of the acidand salt,respecitvelyusedto preparethe buffersolution Takinglogarithm on both sidesof theequation ` log [ H_(3)O^(+)] = log K_(a) + log ("[acid]")/("[salt]")` reverse the sign on bothsides ` - log [ H_(3)O^(+)]= - log[ H_(3)O^(+)] = - log K_(a) - log (" [ acid]")/("[salt]")` We knowthat ` pH = - log[ H_(3)O^(+)] and pK_(a) = - log K_(a)` ` RightarrowpH = pK_(a) - log ("[ acid"])/("[salt]")` ` RightarrowpH = pK_(a) + log(" [ salt]")/("[acid]")` Similarly fora basicbuffer, ` pOH =pK_(a) + log("[ salt]")/("[ base"]))` (ii) Kohlraush's LAW : At infinite dilution , THWE limitingmolarconductivityof anelectrolyte is equalto the sam e of thelimitingmolarconductivites of itsconstituent ions. |
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