You need to prepare 100.0 ml of a ph 4.00 buffer solution using 0.100 m benzoic acid (pka = 4.20) and 0.120 m sodium benzoate. how many milliliters of each solution should be mixed to prepare this buffer?

First, we have to use Henderson-Hasselblach equation to get the weak base/weak acid ratio:

PH = Pka + ㏒[A]/[HA]

when PH = 4

Pka = 4.2

A is the benzoate

HA is the benzoic acid

so,

4 = 4.2 + ㏒[A]/[HA]

∴[A]/[HA] = 0.631

when [A]/[HA] = 0.12 m * VA / 0.1 m * VHA

∴0.631 = 0.12m * VA / 0.1m * VHA

now, when we have the total volume of the solution is 100 mL

∴VA = 100 mL - VHA (we can substitute with 100 - VHA instead of VA)

∴0.631 = 0.12 * (100-VHA) / 0.1 * VHA

we can assume that VHA = X

so,

0.631 = 0.12 * (100-X) / 0.1 * X by solving this equation for X

∴X = 65.5

∴VHA = 65.5 mL

so, VA = 100mL - 65.5 ML

= 34.5 mL