Question

A 50.0 mL solution of 0.112 M KOH is titrated with 0.224 M HCl . Calculate the pH of the solution after the addition of each of the given amounts of HCl . 0.00 mL pH = 5.00 mL pH = 12.5 mL pH = 18.0 mL pH = 24.0 mL pH = 25.0 mL pH = 26.0 mL pH = 31.0 mL

A solution of 112 M KOH ist of the given nousis of HCL with 0.24 MHL Cake the pile the r e is on 0.00 mL -13.05 3.00 13.00 12.5L pH- 240 pilt 250ml. pl

Answer 1

| Кон t не ) ko +н, о 50ml 0.112M 0.224 M r @ poha Millimales of KOH = 50 x 0.11 2 = 5 6 8 10-3 molen o.ome Hee = 5.6 1000 - 50 - 10q [ooll2 J . PH-14-polt=_13.049, in HL - 50 mL Setro en 0.224 x 5.0 = 1.12 x 103 mole will react with 10 12 x 10-3 mole kon, excess kon = 4.48 X lo moles Total volume = 50+5 = 55mL. Com-] = 4.48X10-3 x103 F0:081 - 109 [0.081] = POH= 1.091 P H = 12-908. Hue- 12.5 mL LYD? Psy 0.22 4 12:5= 2.8x10-3 mole will react with 2.88 10-3 male kon, excess Roll = 2.8 x 10-3 mole. Total volume 50 + 12 5 62.5 mL Orlato con-] 2.8 X10-3 x 103. 0.0448. porn 62.5 baş plon] = -log[0.0 448] = 10 3487 pH = 12.6513 He = 18.0mL 0.224 x18 = 4.032 x 10-3 måle. excey kome (506x10-3) -( 4.032 x103) = 1.568 x 10-3 mole - Petad volume i 50 +18=68mL con: 1:56 8 x10-38103. 0:0 230 68 plony = -log[0.0230}= 1.638 H = 12.3617

classmate Date Page Hae: 250 ml 0.224 x 25-0 = 5-6 X10-3 mole Hll. kon=5.68 los mole. Roth will neutralize each other. pH = 7. HU = 26.0mL. 0.224 x 26-0=5.824 x 10-3 mole Hll. kon= 5.6*10-3 mole kon lexces Hee = 0.22 4x10-3 mole. In que Total volume = 50+ 26= 76 mL Chl= 0.224 8 10 3 x 103 = 0.00 29 47 76 - 2 - -| (o' 0 029 449 De 2.5305 Hel=31.0 mL 0.22 4 8 31 = 6.944 8 10-3 mole non kon= 5.6.8 10-3 mole.. 30 y гжсем не : +39 Чx to me : 1 Total volume = 50+3 12 8 l m Lantas a CH+ = 1340 r1o-3 x103 81 0.016592 . ph = -log(0.16592) - = 1.78. 1-78. A