MCQ
The equivalent weight of $H_3PO_4$ in following reaction is $H_3PO_4 + Ca(OH)_2 \to CaHPO_4 + 2H_2O$
- A$98$
- ✓$49$
- C$32.66$
- D$40$
$H_3PO_4 + Ca(OH)_2 \to CaHPO_4 + 2H_2O$
In this reaction
$H_{3} P O_{4} \rightarrow H P O_{4}^{2-}+2 H^{+}$
Orthophosphoric acid leaves $2$ hydrogen ions.
Hence, its valency factor is $=2$
Equivalent weight $=\frac{\text { Molecular Weight }}{2}$
Molecular weight $=1 \cdot 3+31+4 \cdot 16=98$
Therefore,
The equivalent weight is $=\frac{98}{2}=49$
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