Question

Problem 4: Given: 1.0 M HOAC (K. = 1.8 x 10) 1.0 M NaOH stock solution Make 500 mL of 0.50 M HOAc/OAc buffer with pH = 5.00

Answer 1

Answers problem.4 . given, 1.0M HOA (Ka= 1.8x10-5) : 1.0M NaOH stock solution. The buffer will be of weak acid HOAe and its salt NGOA E COA ). det, us suppose, (v) volume(m2) of wom HOAC is mixed with (V2) mL, volume oy 1.0 NaOH.(V. XV 2) W, V2 mL of 1.0M HOAC will be converted salt, in the nei :(V-V2) mL o 7 HOM HOA Memained tom . [one@] Ve [ HOA e] (Vi-v2). Oit, [HOAė] VI-V₂ Tong = V W I according henderson equation: pH = pka - log [Home 7 5= -log (1-8810-5) - log [Home] с. Голее) 7 5 = 4.74 – log LOACET → 0.26 = -log [Home] [DA ] . (HOACT ToAce .: [HD] = cos vvz.0.55 Va VI-V2 = 0.552 on, v= 0.55V2 + ₂ (M= 1.55 V2]. again, Total volume of mixture to be 250mL (250ml 1.0 M (110AC)/(onej produced, then add 250 ml water, to get 10.5M THOAI/1040] ) OAC . 10-0. 26 10 - 256

:: Vit V2 = 250mL. or 1.55V2 +V2 I 250 mL 2.55 V2 = 250 ml :: V2 = 250 mL/2.55 = 98 ml. ..; v = (250–98) = 152 ml. Hence, 152mL of 10 m [Home] added to 98 m2. Ofrom NaOH gives, 250mL of 1.0M [Hote]/[oroj buffered at 5H= 5. Now, add 250 mL q water to the system, then it will become soomL of oism [Hone] /[01e@] buffered at pH=5;