Question

You have been given the task to analyze a large number of samples that contain an unknown concentration of the diprotic acid, hydroselenic acid. You know the concentration is approximately 0.05 to 0.10 M H,Se. However, you do not want to complete a myriad of pH titrations, so you decide to identify some of the expected pH. alues you will encounter. In order to do so. vou have prepared 50.00 mL of 0.105 M H, Se solution as your test solution. You will be titrating the H, Se solutions with 0.180 M KOH. Based on this information: (1) Calculate the pH of the titration solution once 28.00 mL of 0.180 M KOH has been added to the 50.00 mL of 0.105 MH,Se. 78.00mL (2) Calculate the pH of the titration solution once the first equivalence point has been reached (3) Calculate the pH of the titration solution once the second equivalence point has been reached. (4) Calculate the pH of the titration solution once 150.0 mL of the KOH has been added. Helpful Information for H, Se: pKg = 3.89 and pKq2 = 11.00 Effective lonic Diameters: HSe = 0.55 nm Se2- = 0.45 nm

Answer 1

- So oos my un pills to be given. Soutin H₂O o 0 1t₂ se tkou kise + mily to o. losx50 0.18 x 28 o . -5.25 mnt 5.04mmil tat 0.21 mm o 5.04 it makes buffer 5.04 so pha pka, tly (salt) = 3.89 + ln (s.oy ) CAdid I = 389 +1.38 pn = 5.27 if ist equivalence pt. occur - kHse forms only which is a amphiprotic aid. PH=pka, + pkan (for amphiprtic acid] ph= 3.84 +11 = 7.445 2

② st and equivalence point only kg se occurs so it can have hydrolyn's only Kyse +420 2 .0 t on- PH= 7+ & Cpka + lg 1) = 7+ 1 [ll+ly (o.los)] pH = 12.01 i 180 ml kou in added so n = 0.18 x 150 H₂Se takon 5.25m 27 mm to K₂ Se + 2H₂O o o to 5.25 10.5 27-10.5 16.5 mm. cond = 165 = 165 = 0.0625 150+ 50 pon = -ls Co. og 25] = 1.08 pu = 14-1.08 pn = 12.92