Question

26) Calculate the pH during the titration of 25.0 ml of 0.500M HCl with a 0.250M NaOH solution at the following point during the titration. millimoles acid initially 12.5 a) 0.00ml of NaOH added Millimoles base added= 0 millimoles acid left = 12.5 or millimoles of base in excess pH = .30 o b) 15.0 ml of NaOH added Millimoles base added = 3.75 millimoles acid left = 8.75 or millimoles of base in excess معاملے = pH f base in excess o 346 093 1.05 c)50.0ml NaOH added Millimoles base added = millimoles acid left = 12.5 0 or millimoles of base in excess pH = 2 d) 65.00ml NaOH added. Millimoles base added = 16.25 millimoles acid left or millimoles of base in excess pH- 1.4 a poll st 12.60 3.75 S. 2220

Answer 1

Given datai Vhcl = 25.0mL MHO=0.SOOM MNOoh = 0. 250M mores of Hal initially = VAUXMHU = 250 x0.500 1000 -0.0125 mol milimores acid initially = 12.5 (a) o.oome of Nowh added mihimoles based added= 0 milimoles acid lelf = 12.5 er milimores of base in excess to PHS0-30 PHE-log CH+] --log(0.500) = 0.30 (6) 15.ome of Naor added. se added = MNaoH X 15.0 -0.250 XIS miunoter of base added 3.75. since, 3.75 milimoles of base is added so, it will neupalit the 3.75 milimoles of acid so, milimoles of oud left = 12.5-3.75 = 8.75 pH = 0.66 since, only 8.75 milimores of acid is left so, It will girera 8.75 milimores of Ht Total volume = VHolt Nath = 25mL + 15 mL=40mL [H +7= B 0.21875 pH = -log(H+] = tog 10.21875) = 0.66

Date @ 50.omi o mach added milimotes a base odded : MNOOH * V Nash 0.250 x milimoles of bose addest 12.5 Since, 12.5 milimores of Nowh will be fully be neutralised by 12.5 milimores of riel So, pH = 78 milimoles 65.come of Naoh added. base added = M NaoH X 65 0 0.250 x 65 milimoles of base added - 16.25 As the milimores of based added is more than the milimores of net so Hel aiuse completely neutralised and milimores of hare wilube in excess. milmores of base in excess = 16.25-12.5 = 3.75 PH = 12062 Here base is in exces. milimores of excess NooH = 3.75. Total volume = VHCl + VNash 250L+65m. - gomL, TOH-] = 3.75 -0.0416 poh = tog forts-tog 10-0416) -438 PH= 14-POH = 14-1:38 = 12.62 Hence, pH = 12062