calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate

Question

2 years ago

11

buffer with ph 4.76. assume that the pka for the dissociation of acetic acid 4.76. how much of each must be added if the desired ph is 5.76?

1 answer:

kirza4 [7] · Answer 1 · 2024-02-16T01:57:50+03:00

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

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