In the following reaction, the equivalent - factor of $A l (H_{2} P O_{4})_{2} (O H)$ is:
$H_{3} P O_{4} + A l (O H)_{3} \to A l (H_{2} P O_{4})_{2} (O H) + H_{2} O$

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Solution

$H_{3} P O_{4} + A l (O H)_{3} \to A l (H_{2} P O_{4})_{2} (O H) + H_{2} O$
$A l (H_{2} P O_{4})_{2} (O H)$ is formed from $A l (O H)_{3}$ which has 3 $O H^{-}$ ions. 2 of these ions get removed to form $A l (H_{2} P O_{4})_{2} (O H)$ , hence the equivalent factor must be 2.

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