Question

For a buffer made from sodium acetate and acetic acid, the K_a of acetic acid is 1.8E-5.
What mass of sodium acetate (NaCH₃CO₂) must be added to 2.50 L of 0.68 M acetic acid to make a buffer solution with pH = 5.75? The answer should be in two significant figures.

Answer 1

Answery Here given, ka(CH₃ COOH) = 118x105 spka = -log ka = -log(1'8x105) = 4.14 pH = 5.75 and [ CH₂ 100H] = 0.68M Henderson- Hasselbaich education can be written as follows : - - pH = pka + log I bore] ..... [acid AD As the given solution contain an aud (CH3200H) and it's conjugate base (CH3 coora), so abore equation can be applied for this butter solution, Now putting the above values in equation I we get , 5.75 = 4.74 +10g [e Hz.coo ora] 0168 [CH3 corona] — = 5.75 -4.74 = 1001 0.68 A, [Chs coora] . JO101 - 10.232 0.68 ar, [etz coora] = (10.232 x 0.68)M= 6.95776 M We know, molarity (M) = No of moles of solute volume of the solution in litre ar 6.95776 n = No of moles of CH coora 2.50 L 32 232 X No of moles of CH₃Coona = 6.95776m x 2.5oL = 6.95776 moles/4 X 2.50 4 1. = 17.3944 moles he know, - Mass of a substance = No of moles of the substance X Molar mass of substance . Mass of CH₂ coora = no of moles of CH₃COONax Molar mass of ch coora = 14.3944 moles X 82.0343 g Imoles = 1426.9379 y 1426.94 g (upto 2 sigrigiont figure) Mass of CH₃coona = 1426.94

Answer 2