Question

1.) A 26.7 mL sample of a 0.495 M aqeuous hydrofluoric acid solution is titrated with a 0.334 M aqeuous sodium hydroxide solution. what is the pH at the start of the titration, before any sodium hydroxide has been added?

pH = ????

2.) what is the pH at the equivalence point in the titration of a 18.3 mL sample of a 0.429 M aqueous hypochlorous acid solution with a 0.479 M aqueous barium hydroxide solution?

pH = ???

Answer 1

1 26.7 ml sample of a 0.495m Hydrofluoric acid solution &s titiated with a 0.334M aqueous Naot solution. What es pH at the start of the titration before any NaOH Gs added ? So we need to considea 26.7 ml of 6-45 М НЕ . THI] = THE

H – 109,0 (0.yah) = 0.305394 pH = 0.3053940 2) 18.3 ml of 0.429M aqueous hypochisions aqueous aced solution with 0.479 M Barrum hydroxide solution. At equivalence point My = Ma va The volume of Ba(OH) can be determine 18.3 x 0.429 = V2 x 0.479 2 = 16.389 mi of Ba(OH)2 Number of mots = volume & molarity on Poster 2 78507 2 16.389x108x0.479 27.8501 x 10 No. of molks of Ba(OH), Hool .

The reaction between Hoa and Balott. 2 Hod & Balot) - 2H2O + Back, Imole So 2 moks of Hoce requires, Balo), 4 Anyting reagext to Berlon We have 7.8507x10 mokes of Hod and Balo H So the limiting reagent is Hoa. 2 molks I moles 7.8507 810" molesta 7.850 7 X 103 3.92535 x 10 moles of Ba(OH). So we have Balott) mencess. The no of molks of Ba(OH), there to the solution after the reaction Trished = 7.8507x03 - 3.92535 x 10

- 3.92535 x 103 molks Now one thing about acids and bases es that their sons desso cate o solution. Ba(OH) 2000 & Bad That means I mole of Balott), desso dates to form a moks of No. of moles of Otto = 2x0.00392535 OH roons. = 0.007850T Total Volume = 18.3 + 16.389 = 34.689 m . The concentration of oto = 0.0078507 0.034689 L SoHo] = 0.2263166 pott = – log (0.2263 166) 2 0.6452833 pH + polt=140 pH = 13.354 T166n.