What mass of sodium benzoate should be added to 150.0 ml of a 0.15 m benzoic acid solution in order to obtain a buffer with a ph of 4.25? (assume no volume change.) ?

When we know the Pka value = 4.19

so, we are going to use H-H equation:

PH = Pka + ㏒[benzoate/benzoic acid]

when we have PH = 4.25

and Pka = 4.19

so by substitution:

4.25 = 4.19 + ㏒[benzaoate / benzoic acid]

∴ [benzaoate/benzoic acid] = 1.148 M

when the [benzoic acid ] = 0.15 m

∴ [benzaoate] = 1.148M * 0.15m

= 0.1722 M

∴ moles of sodium benzoate = molarity * volume L

= 0.1722 M * 0.15 L

= 0.02583 moles

∴ the mass = moles * molar mass

= 0.02583 moles * 144 g/mol

= 3.72 g