Ka = 4.44*10^-6
pKa = - log (Ka)
= - log(4.44*10^-6)
= 5.353
use formula for buffer
pH = pKa + log ([LiA]/[HA])
3.08 = 5.3526 + log ([LiA]/[HA])
log ([LiA]/[HA]) = -2.2726
[LiA]/4.064 = 0.0053
[LiA] = 0.0217
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 2.169*10^-2*1
= 2.169*10^-2 mol
Answer: 2.17*10^-2 mol
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