
If the equivalent weight of \[{{H}_{3}}P{{O}_{4}}\]in the reaction is \[x\,g\], then find the value of \[x\].
\[Ca{{\left( OH \right)}_{2}}+{{H}_{2}}P{{O}_{4}}\to CaHP{{O}_{4}}+2{{H}_{2}}O\]
A. 49
B. 39
C. 40
D. None of the above
Answer
575.1k+ views
Hint: An equivalent weight of an answer is characterized as the molecular weight of the solute divided by the valence of the solute. Equivalent weight is utilized for anticipating the mass of a substance that responds with one atom of hydrogen in a corrosive base investigation like in titration.
Complete step by step solution:
The given reaction is:
\[Ca{{\left( OH \right)}_{2}}+{{H}_{2}}P{{O}_{4}}\to CaHP{{O}_{4}}+2{{H}_{2}}O\]
In the reaction, \[{{H}_{3}}P{{O}_{4}}\to HPO_{4}^{2-}+2{{H}^{+}}\] 2 hydrogen ions dissociate from the orthophosphoric acid.
Therefore, the valency factor \[=2\]
Equivalent weight \[=\dfrac{molecular\, weight}{2}\]
Molecular weight calculation.
Number of \[H\] is 3
\[3\times 1=3\]
Number of \[P\] is 1
\[1\times 31=31\]
Number of \[O\] are 4
\[4\times 16=64\]
Molecular weight \[=3+31+64\]
Molecular weight \[=98\]
Number of hydrogen ions replaced by the base of the acid is 2.
Therefore, the equivalent weight \[=\left( 98/2 \right)=49\]
Equivalent weight of \[{{H}_{3}}P{{O}_{4}}\] is 49.
Hence, the correct option is A. 49
Additional Information:
The equivalent weight of an element or radical is equivalent to its atomic weight or formula weight isolated by the valence it accepts in compounds. The unit of equivalent weight is the atomic mass unit; the amount of a substance in grams mathematically equivalent to the equal weight is known as a gram equivalent.
Note:
We must know the proper concept of searching the valence factor for acid and base. Acids based on phosphorous like \[{{H}_{3}}P{{O}_{4}}\](phosphoric acid), \[{{H}_{3}}P{{O}_{3}}\](phosphorous acid), \[{{H}_{3}}P{{O}_{2}}\](hypo-phosphorous acid) need specific kind of attention to understand their basicity. Formula depicts that these compounds might furnish \[-OH\] ions, but bond strength between the phosphorus and oxygen \[\left( P-O \right)\] and phosphorus and hydrogen \[\left( P-H \right)\] are stronger than between oxygen and hydrogen \[\left( O-H \right)\].
Complete step by step solution:
The given reaction is:
\[Ca{{\left( OH \right)}_{2}}+{{H}_{2}}P{{O}_{4}}\to CaHP{{O}_{4}}+2{{H}_{2}}O\]
In the reaction, \[{{H}_{3}}P{{O}_{4}}\to HPO_{4}^{2-}+2{{H}^{+}}\] 2 hydrogen ions dissociate from the orthophosphoric acid.
Therefore, the valency factor \[=2\]
Equivalent weight \[=\dfrac{molecular\, weight}{2}\]
Molecular weight calculation.
Number of \[H\] is 3
\[3\times 1=3\]
Number of \[P\] is 1
\[1\times 31=31\]
Number of \[O\] are 4
\[4\times 16=64\]
Molecular weight \[=3+31+64\]
Molecular weight \[=98\]
Number of hydrogen ions replaced by the base of the acid is 2.
Therefore, the equivalent weight \[=\left( 98/2 \right)=49\]
Equivalent weight of \[{{H}_{3}}P{{O}_{4}}\] is 49.
Hence, the correct option is A. 49
Additional Information:
The equivalent weight of an element or radical is equivalent to its atomic weight or formula weight isolated by the valence it accepts in compounds. The unit of equivalent weight is the atomic mass unit; the amount of a substance in grams mathematically equivalent to the equal weight is known as a gram equivalent.
Note:
We must know the proper concept of searching the valence factor for acid and base. Acids based on phosphorous like \[{{H}_{3}}P{{O}_{4}}\](phosphoric acid), \[{{H}_{3}}P{{O}_{3}}\](phosphorous acid), \[{{H}_{3}}P{{O}_{2}}\](hypo-phosphorous acid) need specific kind of attention to understand their basicity. Formula depicts that these compounds might furnish \[-OH\] ions, but bond strength between the phosphorus and oxygen \[\left( P-O \right)\] and phosphorus and hydrogen \[\left( P-H \right)\] are stronger than between oxygen and hydrogen \[\left( O-H \right)\].
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