Question

Determine the pH during the titration of 26.2 mL of 0.296 M perchloric acid by 0.200 M barium hydroxideat the following points:

(1) Before the addition of any barium hydroxide

(2) After the addition of 9.70 mL of barium hydroxide

(3) At the equivalence point

(4) After adding 25.2 mL of barium hydroxide

Answer 1

as given: volume of perchloric acid =26. 2mL = 26.28103 L Molarity of perch logic acid (MC104) = 0.296M Moles of 44104 = Molarity of Hel0qx - volume of nelog = 0.296 mol x 26-2x1024 = 0.0075552mol Molavity of Bacon), = 0.200M (12 volume of Bacon),-0 ML Helog ut +4104 Thus, [uclop] = Cut] = 0. 296M PH = -log (ut) = log (6.296) = 0.528708289 = 0.529

22 volume of Bacon), = 9.70 ML = 9.70 x16 L Moles of Bacon? = Molarity of Bacon2x rolume of Bacon 0.200 mol x 9.70 xIBK = 0.00194 mol Total volume = volume of HC1047 Volume of Bacon) = 26.2 mL + 9.70mL = 35.9mL = 35.9x10?L Excess moles of holoq = Moles g 4cloq- Moles of Ba(OH)₂ = 0.0077552 mol - 0.00194 mol = 0.0058152 mol (nt) Excess moles g KC104 Total volume = 0.0058152 mol 35.9x103 L = 0.161983287M.

- pH = -log Cht] = -log(0.161983287) = 0.790529792 = 0.791

(3) since Bacon is strong base and H. 104, a Strong add, pH at equivalence point = 7.00 (heutral) 643 volume of Bacoma = 25.2mL =25.2x1024 Moles of Bacon = Molarity of Baconax volume of Bacon = 0.200 mallex 25.241034 = 0.00504 mol Total rolume = volume of helpt rolume of Bacon) = 26.2m2+ 25.2mL = 51.4 ML = 51.4D0L Excerse moles of H2104 = Moles of H4104 - Moles of Bacolo = 0.0077552 mol - Oro0504 mol -0.0027152 mal [ut] = Excess moles of Reloa Total rolume =0.0027152 mal 20.0528249027M 57.4x1o3L

PH = -log[ut] = -log (0.0628 24 9027) = 1.27756129 = 1.28