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

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.