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

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

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