Question

40.0 ml of 0.100 M HCl is titrated with 0.100 M KOH. Calculate the pH in the titration OM KOH Calculate the the a t each of the following steps a) Initially before any KOH has been added. b) 200 ml of KOH has been added c) 39.0 ml of KOH has been added d) 40.0 ml of KOH has been added e) Sketch the titration curve

Answer 1

Reaction between Hu and KOH Hulap.) + Koncol.) → kullap) + 42010) 40.0mL O ODM DVOM pH of solution is defined as negative legithum of hydrogenion concentration. pH=-log [ut Huis shong awed and dissociate completely pH = -log (0.100) $4 = [1] initially pH before any addition of KOH= 1 initial mol J HU = Molauity X volume (L) (1 ml=0.00127 – 0.1 mol x 0.04% = 0.004 mol Yomi = 0.042) 9 b Irusial molf Kol = 0.1 mol x 0.02x = 0.002mol I mol of kou react with I mol He 0.002 mol kou rest with 0.002 molte so kou is limsing regent so moly nice left = 0.004 mal = 0.002.ma = 0.002 mer Total volyme = 0.04L+0.024 = 0.06L Molan'tua hier Molarity of HQ = 0 0.002mo) -0.033 M 0.06 L

Би + Bolqi sm 2 – 4 Chtj = -led to o33) = [ 1-чи | 9 340 mL1 кои авай. mol и кои add44 = 0-034ұхо• | moj = O •oo39 соој во" О. oo34 mоl а кои Yes to oo34 тој не Again kol is limsing seagent and consume completely so molgne left = 0.004-0.0039 = 0.00о то concentration of hu= d.oool mol о: 019 = 0.001265 M ри = -а ги+] - - log (o-ool265) = 2-8 4 ) d) 40.0 MLJ Ron added О-ооч то] ) Коч мҳt oi o'004 mol a hu до Кой е ту Plelel, neroupe hd• so neapsalisation occur. so pu= 7:0