Question

please solve the calculation page

DATA SHEET: PH, BUFFERS AND K, Part 1: The Titration of a STRONG ACID (HCI) with NaOH using pH meter a) pH of 25 mL distilled water 5.65 (Measure volume with graduated cylinder) b) pH of 25 mL distilled water with 1 drop 1.0 M HCI 2.92 c) pH of 25 mL distilled water with 10 drops 1.0 M HCI (Note: (H') in 2c is about 10 times that of 2b) 1.10 d) Titration of solution c) with 1.0 M NaOH. 1 M NaOH added drops 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 [(obs) 190/2.04/2. 16/230 2.60 3 13.90 3.40 4.50 1120 wiso 11 lag 1 INSTRUCTOR'S APPROVAL: Part 2: Titration of a WEAK ACID (Acetic acid) with NaOH a) pH of 25 mL distilled water 5.68 b) pH of 25 mL distilled water with 1 drop 1.0 M Acetic Acid. c) pH of 25 mL distilled water with 10 drops 1.0 M Acetic Acid. 3,36 d) Titration of solution e) with 1.0 M NaOH. 1 M NaOH added 12 13 14 15 drops 1 2 3 4 5 6 7 8 9 10 11 Pub 3.3.54|4/09/13747515.05 5.55.546.596.7 10:30 INSTRUCTOR'S APPROVAL:

Answer 1

The pH ralues it has the difference of both cold are not closely associated, of pH 1. The pH of ca from distilled water is 5.8 the outside atmosphere. due to the absorption of TTT one aup of Hol is equivalent o, osmL ic. SxorL. Tray - I Mitel X 5x10th a sx 103 mol HCl ( 6 L - The initial amcentration of HCl is corral to the number of moles of Hil added to the beaker divided by the volume of water to which the Hll was added. 5x105 mol HCl - - 2x103 MHCI 0.025L so this 2x10 3n HCl dissociates completely to give (HD in pH2 -log (H₂O) = -log (2x155) PH= 2.699 - I drops ad dihim of HCl to some of water. (themitical) But obtained is pH= 2.92 . The difference is = 2.92- 2.699 ApH = 0.221

Therritical pH after the addition 25 m2 of distilled water, of 10 drops of Hol to - IN HCL x 10x Sx10TL = 5x164 moles - 5x104 moles = 0.02m Hel. يا أي(afx Het pHz - log (mg] = -log (0.036) (pH = 1.698] Theortical pH [PH = ~1.90 obtained pH ez dilexenre is = 0.202. in case 16 it is Hol (6) Because a stong acid dissociates complethy and accounts for the low pH value. But in case of 20 dissociates partially it is acetic acid a weak acid this accounts for the difference in the pH. (6) As you know the acetic acid is weak acid and dissociate partially since the dissacralm is very weak hence low arailability of (+1] Ims hence even ater adding 4-6 drops of Nach pH does not aller much. CH₃COOH + NaOH - CH₃COONa+ 40.

(d) 36) 1. Because the the change in butter resists acid. pH after adding base e Assume weak acid as HA HA - Naoll - eur) HNa + ay) ĻOcas (10) - As you know the bulter is one thing which resists the change in pH even after adding acid and bases but change ofter adding appreciable amount of acid occurs added. base is - As per the change in pH at half neutralisatm is 5.23 ie pH=pka. The forthor addition of NaOH accounts for the least change in pH. & Assume Salt ar MH 40 MHI - Haw MOH + H₂J 4B) MH + HCl the tid + MH ann a91 pHis accented by [ted] (46) MHI NAOH HOL) Mihaw + oit + Nat. pH ir accounted

1 40) M Hawr + NaOllan + Hon Ml + 14tt Nattoo - Ho - The pH is accounted for ons ② Molecular weight of the Acid unknown = 9 Giren. =) volume of acid = 29.50ml. & molarity 2 0.1988. neight of the acid = 1.1681 g Weight of acid molecular norght = molecular weight molarity x wolumex 1000m2 1,168 8 = 0-1988 mx 29.50ml x molecularmish 1 roomL molecular weight = 198.57 ② pH of half neutralised i pHapka= 5.23 point is equal to pka. (from 39. 2.) rat pla= -log[ka] 5.232-log (ka] (kaz 5.8881106

pH of richinal acid solutim = 2.94. ka= 5.888x156 pH = -loy (HI (10) = -2.94 (CHU = 1.148 x163 - HA+HO Z H + Á ka= Ho] [A] [HA] 5. 8x106 = 1.1488103 x 1.148x103 (14] [[had = 0.223 M 6 85ml x 0.2239 L ) 25x 1034 x 0.223 M molarity = 5.5 75x10 mols.