Question

33.0 mL of 0.170 M KN3 is titrated with 0.250 M HCl. acid-base table a) What volume of acid is required to reach the equivalence point? volume of acid = mL b) What is the pH of the solution after addition of 11.6 mL of acid? pH = c) What is the pH of the solution after addition of 24.2 mL of acid? pH = d) What is the pH at the equivalence point? pH =

Answer 1

2 KN + He kut Hng. and Bare (1) Mi via M2V? 9 0:170x 33.270.250.X X 0: 170 X 33 .. 0.250 2 2244 me of Hill: (b) 11. 6.me of and is added then concot (ht...? MiVi= M2 Ve 0:170 33 = x x 11.6 x 0,1 Ż0 x 33 NL 1106 .0.48362 - (af? 12 So pole - log fut] 1 ) canned with Ph = -log [ory 83627 'CamScanner - 0:3155

24.2 me of and added Miv 2 M2 V2 MX 24. 2 0.170 X 33 - -) XC 70:170 X 33 24.2 1931 M 0.231 so cene of fit = 0.231 = 2log fatla py Pha Llog [o: 231) - 0.639 1 = (d) Phat equivalence point 33.me of 0.170 M KN will be turated with 0.250m of Hill 2004 = 22.44 ml cs Scanned with CamScanner

aut equuvekleme No olmoles of thei: 0.250 x xiure . 22 / 4 x = d.250 x 22:44 lorox Ovousz 1 moles » e No of moties of kn Xurs 0:170 m 0.170 .331 ; H = W tice lo.. ino.rusel | Noof moles of kNo = No of moles of at equivalence so KN₂ + nu → kut HN₃. E Stänner a : a d o ..camScannor ö a la

So aften equivalence the cone of this same as inc of HN that is equivalent to one of hee 120.200 M SolPH = 18-1rg HT = -log [o wo day 0 -60 2050 Scanned with CamScanner