Question

The titration curve with 4 different points is given as well. I am not sure, do I just need to obtain the values of the pH for each point from the graph or I need to calculate something?

1. Calculate the pH of a 0.200 M HCH,O2 solution. Titration of 25.00 mL of 0.200 M HC2H2O2 with 0.200 M NaOH 14 Hd Volume (mL) of 0.200 M NaOH

Answer 1

of 25mL of 0. 200 (my HC₂H30, with 0.200 m Titration NOOH : a lot DH 88 4 B . 0 10 20 30 40 50 Vol my of 0-2(M) NaOH Ans: - You can determine the value of pH at each point from the graph and also you can calculate the value of pH by using formula At point A: ÞH = pka - 5log [acid] Here, bka = -logka, ka = Dissociation constant of CH3 200M Lacid] = Cone. of Chts 2004 At point B: pH = pka : At point < ? pH = { pkw. + { pka + { 107 [sast] Pkw = -log kw , Kw = gonic product of water [salt] = Conc. of Eltz COO NG formed btring by at equivalance point. At point D : pH = + Kw + long [OH-] 3 [OH-] = Cons of NaOH aut point de

By using these formula. you can calculate. the pH at those a different points . A The pka of .CH3COOH = 4.8. Now, At point A: DH = ļ pka - { log [acid] i ÞH = } *4.8 - log (0:2) Lacid] =0.2M I giren , = 2:4 + 0.349 54. = 2.79] at point A. At point B: PH = pka = 4.8 At point cibH = Pkw + PKq + < log Esalt] squivalance point) [salt] = 2550*2 = 02 50.1 At equivalance point, 125 mL acid + 25 ml base) is present, soi total volume = 50 m2 :. pH = 3x19 + Źx4-8 + Ź log (0-0) = 7+ 2.4 + (-0.5) = 8.9 lat point D At points : pH = 6 pkw + log [ou] totaj yolume panir Here, [OH-] = 0-2X10 = 0057 . : [H] = to evolution 5175X10"} pH --- 109 [**] = 12.75 Or, pH= 14 + log [OH-] = 14 + log (0-057) = 12.75 10-14 10-14 LOH] 175 x 103 0.157

By using Formula you can calculate Hoe value Of or pH, then you can cross-check whether your calculation is right or wrong from the graph. 1 In the question written calculate the PH, so i think you have to calculate the value of pH by using the formula