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?

The volume of 0.100 M benzoic acid required is 65.6 mL and the volume of 0.120 M sodium benzoate is 34.4 mL.

Buffer solutions

Buffer solutions are solutions that resists changes in its pH when a little quantity of acid or base is added to it them.

The Henderson-Hasselblach equation

The Henderson-Hasselblach equation is used to determine the weak base/weak acid ratio in a buffer solution

Henderson-Hasselblach equation is given as: pH = Pka + ㏒[A]/[HA]

From the data provided:

pH of buffer = 4.00

pKa of acid = 4.20

A is benzoate and benzoic acid is HA

Using the Henderson-Hasselbach equation:

4.00 = 4.20 + ㏒[A]/[HA]

[A]/[HA] = 0.63

Also; Molarity = number of moles × volume

[A] = 0.12 × V1

[HA] = 0.10 × V2

Hence:

0.1 2 × V1 / 0.10 × V2 = 0.63

V1 = 0.10 × 0.63 × V2 / 0.12

V1 = 0.525 × V2 eqn 1

Also, since V1 + V2 = 100 mL

V2 = 100 - V1

Substitute V2 = 100 - V1 in eqn 1

V1 = 0.525 × {100 - V1}

V1 = 52.5 - 0.525V1

1.525V1 = 75.6

V1 = 52.5 ÷ 1.525

V1 = 34.4 mL

Then V2 = 100 - 34.4

V2 = 65.6

Therefore, the volume of 0.100 M benzoic acid required is 65.6 mL and the volume of 0.120 M sodium benzoate is 34.4 mL.

Learn more about buffers and Henderson-Hasselblach equation at: link

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
 

Hno3(aq) → h+(aq) + no3−(aq). picture ii best represents the strong acid hno3. strong acids are strong electrolytes

High-energy and unstable.