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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