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 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 is used to determine the weak base/weak acid ratio in a buffer solution

From the data provided:

pH of buffer = 4.00

pKa of acid = 4.20

Using the Henderson-Hasselbach equation:

4.00 = 4.20 + ㏒[A]/[HA]

[A]/[HA] = 0.63

[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

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> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.