You have to prepare a pH 3.65 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many mL of HCOOH and HCOONa would you use to make approximately a liter of the buffer?

550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M

Explanation:

It is possible to find the pH of a buffer by using H-H equation:

pH = pKa + log [A⁻]/[HA]

For the formic buffer (HCOOH/HCOONa):

pH = 3.74 + log [HCOONa]/[HCOOH]

As you need a buffer of pH 3.65:

pH = 3.74 + log [HCOONa]/[HCOOH]

3.65 = 3.74 + log [HCOONa]/[HCOOH]

Where  [HCOONa]/[HCOOH] can be taken as the moles of each specie.

As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:

Replacing (2) in (1):

0.81283 = 0.10 - [HCOOH]  /[HCOOH]

0.81283[HCOOH] = 0.10 - [HCOOH]

1.81283[HCOOH] = 0.10

And moles of HCOONa are:

[HCOONa] = 0.1 mol - 0.055mol =

As concentration of the solutions is 0.1M, the volume you need to add of both solutions is:

HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M

HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

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