In: Chemistry
pka of benzoic acid =
-logka
= -log(6.28 *10^-5)
= 4.2
pka of chlorous acid = -logka
= -log(1.12 *10^-2)
= 1.95
pka of hypobromous acid = -logka
= -log(2.3 *10^-9)
= 8.64
so that, best one to get the desire buffer is benzoicacid
pH of acidic buffer = pka +log(C6H5COO-/C6H5COOH)
pka = 4.2
total no of mol of buffer = 0.25*1 = 0.25 mol
4.9 = 4.2 +log(x/(0.25-x))
x = 0.21
No of mol of C6H5COONa = x = 0.21 mol
amount of C6H5COONa = n*Mwt
= 0.21*144.12
= 30.26 g
No of mol of C6H5COOH = 0.25-0.21 = 0.04 mol
amount of C6H5COOH = n*Mwt
= 0.04*122.12
= 4.9 g