Equivalent weight of $H_{3} P O_{4}$ in each of the reaction will be respectively:

$H_{3} P O_{4} + O H^{-} \to H_{2} P O_{4}^{-} + H_{2} O$
$H_{3} P O_{4} + 2 O H^{-} \to H P O_{4}^{2 -} + 2 H_{2} O$
$H_{3} P O_{4} + 3 O H^{-} \to P O_{4}^{3 -} + 3 H_{2} O$

Question

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Answer 1

$E q u i v a l e n t w e i g h t = \frac{M o l e c u l a r w e i g h t}{N u m b e r o f E q u i v a l e n t s}$
In case of:

$H_{3} P O_{4} + O H^{-} \to H_{2} P O_{4}^{-} + H_{2} O$

$H^{+}$ ions given = 1
Thus the number of equiv $= 1$
$E q u i v a l e n t w e i g h t = \frac{9 8}{1} = 98$

In case of:
$H_{3} P O_{4} + 2 O H^{-} \to H P O_{4}^{2 -} + 2 H_{2} O$
$H^{+}$ ions given = 2
Thus the number of equiv $= 2$
$E q u i v a l e n t w e i g h t = \frac{9 8}{2} = 49$
In case of:
$H_{3} P O_{4} + 3 O H^{-} \to P O_{4}^{3 -} + 3 H_{2} O$
$H^{+}$ ions given = 3
Thus the number of equiv $= 3$
$E q u i v a l e n t w e i g h t = \frac{9 8}{3} = 32.67$

Answer 2

E q u i v a l e n t w e i g h t = \frac{M o l e c u l a r w e i g h t}{N u m b e r o f E q u i v a l e n t s}

In case of:

H_{3} P O_{4} + O H^{-} \to H_{2} P O_{4}^{-} + H_{2} O

H^{+}

ions given = 1

Thus the number of equiv

= 1

E q u i v a l e n t w e i g h t = \frac{9 8}{1} = 98

In case of:

H_{3} P O_{4} + 2 O H^{-} \to H P O_{4}^{2 -} + 2 H_{2} O

H^{+}

ions given = 2

Thus the number of equiv

= 2

E q u i v a l e n t w e i g h t = \frac{9 8}{2} = 49

In case of:

H_{3} P O_{4} + 3 O H^{-} \to P O_{4}^{3 -} + 3 H_{2} O

H^{+}

ions given = 3

Thus the number of equiv

= 3

E q u i v a l e n t w e i g h t = \frac{9 8}{3} = 32.67