Question

Why $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$$ is triprotic acid but $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$$ is diprotic acid?

Answer 1

Hint: Phosphorus forms a number of oxoacids. Each oxoacid has at least one $${\text{P}} - {\text{O}}$$ unit and one $${\text{P}} - {\text{O}} - {\text{H}}$$ group. The $${\text{O}} - {\text{H}}$$OH group is ionisable but the H atom linked directly to P is non-ionisable. Thus, the number of $${\text{O}} - {\text{H}}$$ groups present in these oxoacids decide the basicity of the acid.

Step-by-step answer:
Phosphorus produces a number of oxyacids. Some common oxyacids are $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$$ (orthophosphoric acid) and $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$$ (phosphorus acid).

Since $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$$ is known as triprotic acid and $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$$ is known as diprotic acid. This can be understood on the basis of their structures.

Basicity of the acid is equivalent to the number of hydrogen atoms which can be replaced by a base.

Lewis structure can be made by using the total number of valence electrons needed by all atoms to achieve noble gas configuration.
In $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$$, there are three hydrogen atoms, one phosphorus atom and three oxygen atoms. Thus the total number of valence electrons (V.E) can be calculated as follows:

$${\text{Total}}\;{\text{valence}}\;{\text{electrons}} = \left( {{\text{3}} \times {\text{V}}{\text{.E}}\;{\text{of}}\;{\text{H}}} \right) + \left( {{\text{1}} \times {\text{V}}{\text{.E}}\;{\text{of}}\;{\text{P}}} \right) + \left( {{\text{4}} \times {\text{V}}{\text{.E}}\;{\text{of}}\;{\text{O}}} \right)$$

Since we know that the atomic number of hydrogen is 1 and its electronic configuration is 1. This shows that there is 1 electron in the outermost shell (valence shell). Thus the valence number of hydrogen is 1. The atomic number of sulphur is 15 and its electronic configuration is 2, 8, 5. This shows that there are 5 electrons in the outermost shell. Thus the valence number of phosphorus is 5. The atomic number of oxygen is 8 and its electronic configuration is 2, 6. This shows that there are 6 electrons in the outermost shell. Thus the valence number of oxygen is 6.

Hence the total number of valence electrons in $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$$ can be calculated as:

$$\displaylines{
  {\text{Total}}\;{\text{valence}}\;{\text{electrons}} = \left( {3 \times 1} \right) + \left( {1 \times 5} \right) + \left( {4 \times 6} \right) \cr
   = 3 + 5 + 24 \cr
   = 32 \cr} $$

This means that there are a total 32 valence electrons needed by all atoms to acquire noble gas configuration. On the basis of these valence electrons, Lewis structure can be made. When each atom acquires complete octet, Lewis structure of $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$$ is ready as follows:

It can be observed from the above structure that there are three $${\text{O}} - {\text{H}}$$ bonds. Thus its basicity is 3 and hence $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$$ is a known as triprotic acid.

In $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$$, there are three hydrogen atoms, one phosphorus atom and three oxygen atoms. Thus the total number of valence electrons (V.E) can be calculated as follows:

$${\text{Total}}\;{\text{valence}}\;{\text{electrons}} = \left( {{\text{3}} \times {\text{V}}{\text{.E}}\;{\text{of}}\;{\text{H}}} \right) + \left( {{\text{1}} \times {\text{V}}{\text{.E}}\;{\text{of}}\;{\text{P}}} \right) + \left( {{\text{3}} \times {\text{V}}{\text{.E}}\;{\text{of}}\;{\text{O}}} \right)$$

Hence the total number of valence electrons in $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$$ can be calculated as:

$$\displaylines{
  {\text{Total}}\;{\text{valence}}\;{\text{electrons}} = \left( {3 \times 1} \right) + \left( {1 \times 5} \right) + \left( {3 \times 6} \right) \cr
   = 3 + 5 + 18 \cr
   = 26 \cr} $$

This means that there are a total 26 valence electrons needed by all atoms to acquire noble gas configuration. On the basis of these valence electrons, Lewis structure can be made. When each atom acquires complete octet, Lewis structure of $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$$ is ready as follows:

It can be observed from the above structure that there are two $${\text{O}} - {\text{H}}$$ bonds. Thus its basicity s 2 and hence $${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$$ is a known as diprotic acid.

Note: The basicity of the acid can be determined by making structures of phosphorus oxoacids. If any phosphorus acid comprises of one $${\text{O}} - {\text{H}}$$ bond, then basicity will be 1, if there are two $${\text{O}} - {\text{H}}$$ bonds, then basicity will be 2 and if there are three $${\text{O}} - {\text{H}}$$ bonds, then basicity will be 3.