Question

Determine the pH during the titration of 22.4 mL of 0.136 M hydroiodic acid by 0.222 M sodium hydroxide at the following points:

(1) Before the addition of any sodium hydroxide

(2) After the addition of 6.85 mL of sodium hydroxide

(3) At the equivalence point

(4) After adding 17.4 mL of sodium hydroxide

Answer 1

Ans. Calculation of pH of solution during titration - Volume of hydroíodlia quid = 22.4ml Molarity = 0.136 Molarity of Nook= 0-2 RM (1) Before addition of any Sodium hydroxide- hydroiodic acid is strong acid it is completely dissociated H I caq ht car I Caq) come. - 0.136 - - log (H+) = -log[0.136] -- -(-0.866) [PH = 0.85 (2) After addition of 6.85 ml of sodium hydroxide. Moles of Naol = conc. x vol. cinL) 50.212 MX 6-85 X1034 = 1.5207 x 10-3 moles = 0.85 Moles of HI = 0.136M X 22.4 X105 L = 3.046YX 10-3 moles Moles of HI left un neutralised - 3.0464 zló3.15207x163 = 1.5257x10-3 moles How concentration of HI in resulting solution is concentration of HI = Moles of HI 1:52:57 X10-3 mol Total volume 6.85*10* + 22.4x10°3)2 : 1.5257 1-3 29.25 X10-3 M = 0.0521 M pH = -log (14+) = -log [0.0501] =-(-1.28) = 1.28 pH = 1.287

(3) At equivalence point. HI Caq) + NaOH Caps Hoo (a) + Na I cer) At equivalence point all the moles of HI is neutralised by Nath And there is Hao left which has pH 7. So, at equivalence point the pH of solution is 7 Gin case of strong acid and strong base) (1) After adding 17.4 mL of sodium hydroxide. Moles of NaOH = Conc. x vol Cine) = 0.222M X 17.4 X10 L = 3.8628 X10-3 mole Moles of HI - 0.136 M X 22.4x103 L = 3.0464 X. 1103 moles Moles of Naom left unneutralised - 3.8628x163_ 3.0464x103 = 0.8164 X16 " moles Coric of Naoh in resulting solution - Conc. of Nooh = Moles of Noord - 0.8164x103 moles Totul volume (in L) (170481034 24.48100) = 0.8164x10-3 mol 0.0205 M 39.8 X10-3L Nook is strong base it completely lonises in solution and hence [OH-] = 0.0205 m poH = -log[OH-] = -log[0.0205) = -(-1.699= 1069 pH = 14- poh - 14-169 - 12.31 pH = 12-31]