a) If we give NaOH 100 ml. pKa=3.39
log([A−]/[HA]) = pH-pKa.
log([A−]/[HA])= 3.5-3.39 = 0.11.
[A−]= 0.1 x 0.68 = 0.068.
0.071 + 0.068 = 0.139.
0.139/0.45 = 0.3088 L. = 308.8 ml.
b) No, because Buffer Solution is a water solvent based solution which consists of a mixture containing a weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base.
c) No, because there NaNO2.