Normality of 3 M phosphorus acid $H_{3} {P O}_{3}$ is:

Open in App

Solution

$N o r m a l i t y = M o l a r i t y \times n . f .$
where n.f. is n-factor i.e. acidity of base or basicity of acid
here acid given is $H_{3} P O_{3}$
therefore it’s basicity is 2 (since it can donate two $O H^{-}$ ions)
Hence,
Normality $= 3 \times 2 = 6 N$

Suggest Corrections

Similar questions

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

0.1 M

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

0.3 M

(H_{3} P O_{3})

The normality of 0.3 M phosphorus acid (H₃PO₃) is:

0.3 M

(H_{3} P O_{3})

Join BYJU'S Learning Program