Question

How many liters of 1.0M HCl is needed to make a 1.0L of 0.1M sodium benzoate (NaOAc) buffered solution at pH 5.50?

Ka = 6.46x10^-5 M

Answer 1

Reaction is

NaOAc + HCl AcOH + NaCl

Ka = 6.46*10^-5

pKa = - log 6.46*10^-5 = 4.2

moles of NaOAc = molarity volume = 1.0 0.1 = 0.1

let moles of HCl added = x

BCA table is

	NaOAc	HCl	AcOH
Before	0.1	x	0
change	-x	-x	+x
After	0.1-x	0	x

Henderson -Hasselbalch equation is

pH = pKa + log ( )

or, 5.50 = 4.20 + log ( )

or, log ( ) = 5.50 - 4.20

or, log ( ) = 1.3

or, = 10^1.3

or, = 19.95

or , 0.1 - x = 19.95 x

or, 20.95 x = 0.1

or, x = (0.1/20.95) = 0.0047

now,

moles of HCl added = 0.0047

now, molarity of HCl = (moles/volume)

or, 1.0 = (0.0047/volume)

or, Volume = (0.0047/1.0) = 0.0047 L = (0.00471000) = 4.77 mL

so, volume of 1.0 M HCl needed is 4.77 mL.