You need to produce a buffer solution that has a ph of 5.39. you already have a solution that contains 10. mmol (millimoles) of acetic acid. how many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? the pka of acetic acid is 4.74.

Here we will use Henderson - Hasselblach equation:

sO,

PH = Pka + ㏒[conjugate base / weak acid]

when PH = 5.39 
and Pka = 4.74 

and when the conjugate base is [CH3COO-] and the weak acid is [CH3COOH]

and when the concentration (C) = no.of moles / volume
 
[CH3COO-] / [CH3COOH] = no.of moles acetate/V  * no.of moles acetic / V

∴ [CH3COO-] / [CH3COOH]  = no.of moles acetate / no.of moles acetic
 
∴ PH = Pka + ㏒[no. of moles acetate / no.of moles acetic ] 

by substitution:

5.39 =   4.74  + ㏒[no.of moles acetate / no.of moles acetic]

∴ [no.of moles acetate / no.of moles acetic] = 4.5

when no.of moles acetic = 10mmol

∴no.of moles acetate = 4.5 * 10mmol 
                                    = 45 mmol