
The monobasic acid among the following is-
[A] ${{H}_{3}}P{{O}_{3}}$
[B] ${{H}_{2}}{{S}_{2}}{{O}_{7}}$
[C] ${{H}_{3}}P{{O}_{2}}$
[D] ${{H}_{2}}{{P}_{2}}{{O}_{7}}$
Answer
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HINT: On ionisation with water the acids which produce a single hydronium ion i.e. readily ionises only one proton, are known as monobasic acids. The basicity/acidity of oxo-acids is defined by the number of –OH bonds present in the acid.
Complete step by step solution:
> We know acids produce hydronium ions (${{H}_{3}}{{O}^{+}}$) on ionisation. The number of hydronium ions produced makes it mono, di or tri-basic.
To find out if the given acid is monobasic tribasic or dibasic, we can also look at the number of –OH bonds in the given acid. If the acid contains one, two or three –OH bonds, they are monobasic, dibasic and tribasic respectively. To find out the number of hydroxyl groups present, we need to draw the structures of each of the given compound-

${{H}_{3}}P{{O}_{3}}$, phosphorous acid has two P-OH bonds and one P-O bond.
${{H}_{2}}{{S}_{2}}{{O}_{7}}$, disulphuric acid, also known as oleum has one S-O-S bond and four S-O bonds and two S-OH bonds.

${{H}_{3}}P{{O}_{2}}$, hypo phosphorous acid has one P-O bond and one P-OH bond.

Lastly, ${{H}_{2}}{{P}_{2}}{{O}_{7}}$have one P-O-P bond and two P-O bonds.

> Here, ${{H}_{3}}P{{O}_{3}}$ is not tribasic as the –H atom present is the acidic hydrogen, it can readily ionize 2 protons (two –OH bonds present here) therefore it is a dibasic acid. Similarly, ${{H}_{2}}{{S}_{2}}{{O}_{7}}$ and ${{H}_{2}}{{P}_{2}}{{O}_{7}}$are dibasic and tribasic respectively.
Since ${{H}_{3}}P{{O}_{2}}$has only one proton available to ionize i.e. only one ${{H}^{+}}$can be donated, therefore it is monobasic.
Therefore, option [C] is the correct answer.
Additional information: According to the Bronsted-Lowrey concept, an acid is a substance that can release a proton and a base is a substance that can accept protons.
Note: It is important to remember here that the presence of –H atoms does not define the acidity of oxo-acids. We get the measure of acidity/basicity by the –OH bonds. It is also important to remember the structures of these common compounds as we often miss out the double bonded O atoms and write them as –OH instead.
Complete step by step solution:
> We know acids produce hydronium ions (${{H}_{3}}{{O}^{+}}$) on ionisation. The number of hydronium ions produced makes it mono, di or tri-basic.
To find out if the given acid is monobasic tribasic or dibasic, we can also look at the number of –OH bonds in the given acid. If the acid contains one, two or three –OH bonds, they are monobasic, dibasic and tribasic respectively. To find out the number of hydroxyl groups present, we need to draw the structures of each of the given compound-

${{H}_{3}}P{{O}_{3}}$, phosphorous acid has two P-OH bonds and one P-O bond.
${{H}_{2}}{{S}_{2}}{{O}_{7}}$, disulphuric acid, also known as oleum has one S-O-S bond and four S-O bonds and two S-OH bonds.

${{H}_{3}}P{{O}_{2}}$, hypo phosphorous acid has one P-O bond and one P-OH bond.

Lastly, ${{H}_{2}}{{P}_{2}}{{O}_{7}}$have one P-O-P bond and two P-O bonds.

> Here, ${{H}_{3}}P{{O}_{3}}$ is not tribasic as the –H atom present is the acidic hydrogen, it can readily ionize 2 protons (two –OH bonds present here) therefore it is a dibasic acid. Similarly, ${{H}_{2}}{{S}_{2}}{{O}_{7}}$ and ${{H}_{2}}{{P}_{2}}{{O}_{7}}$are dibasic and tribasic respectively.
Since ${{H}_{3}}P{{O}_{2}}$has only one proton available to ionize i.e. only one ${{H}^{+}}$can be donated, therefore it is monobasic.
Therefore, option [C] is the correct answer.
Additional information: According to the Bronsted-Lowrey concept, an acid is a substance that can release a proton and a base is a substance that can accept protons.
Note: It is important to remember here that the presence of –H atoms does not define the acidity of oxo-acids. We get the measure of acidity/basicity by the –OH bonds. It is also important to remember the structures of these common compounds as we often miss out the double bonded O atoms and write them as –OH instead.
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