You visited us 1 times! Enjoying our articles? Unlock Full Access!

Neutralisation Reaction

Equivalent we...

Question

Equivalent weight of $H_{3} {P O}_{3}$ is:

E = \frac{M}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

E = \frac{M}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

E = \frac{M}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $E = \frac{M}{2}$
Equivalent Weight of $H_{3} P O_{3} = \frac{M o l e c u l a r w e i g h t o f H_{3} P O_{3} (M)}{B a s i c i t y}$
Since, $H_{3} P O_{3}$ is a dibasic acid i.e it has two hydrogen ions to donate to a base in acid - base reaction.
$H_{3} P O_{3} \to 2 H^{+} + H P O_{3}^{-}$
So, equivalent weight of $H_{3} P O_{3} = \frac{M}{2}$
Since, it has a basicity of $2$

Suggest Corrections

Similar questions

m_{A}

g

A

m_{B}

g

B

E_{A}

E_{B}

F e_{0.95} O

\to F e_{2} O_{3}

F e_{0.95} O

I O_{4}^{-}

I_{2}

H_{3} {P O}_{3}

2 S_{2} O_{3}^{2 -} (Reducing ~agent) \to S_{4} O_{6}^{2 -} + 2 e^{-}

S_{2} O_{3}^{2 -}

M

S_{2} O_{3}^{2 -}

Join BYJU'S Learning Program