If K_(2) and K_(2)are theionizationconsatnats of H_(3)NCHRCOOH and H_(3)overset(+)(N)CHRCOO^(-) , respectively , the pH of the solutionat theisoelectric point is .

If K_(2) and K_(2)are theionizationconsatnats of H_(3)NCHRCOOH and H_(

1.	If K_(2) and K_(2)are theionizationconsatnats of H_(3)NCHRCOOH and H_(3)overset(+)(N)CHRCOO^(-) , respectively , the pH of the solutionat theisoelectric point is .
Answer» `pH = pK_(1) + pK_(2)` `pH = (pK_(1) pK_(2))^(1//2)` `pH = (pK_(1) + pK_(2))^(1//2)` `pH = (pK_(1) + pK_(2))//2` Solution :`H_(3) overset(+)(N) UNDERSET("LOW pH")overset(R)overset(\|)(CH)COOH underset(H^(+))overset(OH^(-))(hArr) underset("Intermediate pH")(H_(3)overset(+)(N)overset(R)overset(\|)(CH)COO^(-)) underset(H^(+))overset(OH^(-))hArr H_(2)N underset("HighpH")(H_(2)N overset(R)overset(\|)CH)COO^(-)` `k_(1) =([H_(3)overset(+)NCHRCOO^(-)][H^(+)])/([H_(3) overset(+)NCHROOH]) , K_(2) = ([H_(2)NCHRCOO^(-)][H^(+)])/([H_(3) overset(+)NCHRCOO^(-)])` Thus , `K_(1)K_(2) = ([H_(2)NCHRCOO^(-)][H^(+)]^(2))/([H_(3)overset(+)NCHROOH])` At theisoelection point `[H_(2)NCR RCOO^(-)] = [H_(3)N^(+) CHRCOOH]` `k_(1)k_(2) = [H^(+)]^(2)` `2 log [H^(+)] =log K_(1) + log K_(2)` `RARR ""-2log[H^(+)] = - logK_(1)- log K_(2)` `2pH = pK_(1) + pH_(2) rArr pH= (pHk_(1) + pK_(2))//2`