Suppose that you have 135 mL of a buffer that is 0.360 M in both propanoic acid ( C 2 H 5 COOH ) and its conjugate base ( C 2 H 5 COO − ) . Calculate the maximum volume of 0.240 M HCl that can be added to the buffer before its buffering capacity is lost.

202.50 ml is the maximum volume of 0.240 M HCl that can be added to the buffer before its buffering capacity is lost.

Explanation:

Data given:

volume of buffer = 135 ml or 0.135 litres

molarity of the buffer = 0.360  M

volume of the acid = ?

molarity of the acid = 0.24 M

the number of moles of conjugate base is

M X V = number of moles

0.360 X 0.135

0.0486 moles of conjugate base

The maximum amount of acid added will equal the amount of conjugate base from the buffer.

So, the volume of acid will be calculated by using the formula:

Molarity =

 volume =

              = 0.2025 l

202.50 ml is the volume of acid added.

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.