Question

please answer and explain everything fully

1. Sketch and fully label a titration curve for 0.09660 M base with a K = 2.4 x 10 titrated with 0.1224 M HNO, Calculate start and end point pH values, the equivalence point volume (why can't you calculate the end-point volume?). There are at least ten points of interest on this curve to grade.

Answer 1

Titration of a weak bayang by a strong aced. ut a consider a weak borse Bolt. B011 rst + oH — C 0 0 c(1-x) ca , ca c is the mitial concentration. & be the degreest dnovation ekz = [st] tot). = - ca Caz - Ks = BOH) CH + LOH) = ca kbc

MARRAS TOH) = 0.09660x 2.4x10-6 polt - log to H) = -wg (10,49660 X2.4*10-6). 3.32. pH = 14-pot = 14-3-32 = 10.68 Vacid & Said = Vbane Xlbare. We can not calculate the end point value as the volume of the base is not given tay here. At the end poent the son. contoun only the corresponangsalt. at this point pH of the sohn is due to my brother of the salt. And then pH of the soln in given by pH = 2 pkw / RKB floge ( conc. of salt. = 7+ / g (2.4810-6) - Y, wgc ; on pkb = -logk, - 4.19 Y loge. W wa We con not calculate the value of ac e, sonuntatun of sult formed on the rolurne of the song is not goven here. but by logical though we com say that pH at ear.poont on and point should be less them mitind pH value. on the conc. of and becomes increases ie, the soln becomes and ana hinu pH value drops. this for this kind of titration we get a curve like this vol of HNO₃ →