Homework Answers
given pH = 5
volume = 10 mL
molarity [HA] = 0.1 M
Ka = 1.2 x 10-2
pKa = -log (Ka)
= - log (1.2 x 10-2)
= 1.92
pH = pKa = log( [A-] / 0.1)
[A-] = 0.12M
mole of base = 0.12 M x 100 ml x 1L / 1000 mL
=0.012 mol
mass of base = 0.012 mol x 322.1 g / 1 mol Na2SO4. 10H2O
= 3.87 g
therefore required mass of base is 3.87 g
initial volume is requires volume for buffer solution =100 mL
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