# Neutralization Reaction problem

$\frac x4$ equivalent of $\ce{NaAl(OH)2CO3}$ is neutralized by $x$ equivalent of $\ce{HCl}$

Milliequivalent of $$\ce{HCl}=\mathrm{{160}\times{0.25}\times{1}\times{40}}/100=16$$

$1$ is for one $\ce{H+}$.

Hence milliequivalent of $\ce{NaAl(OH)2CO3}$ required is $4$

$\mathrm{4 = {0.1}\times{V}\times{2}}$

$2$ is for two $\ce{OH-}$ So, $V = \pu{20ml}$

But given answer is \pu{40ml}, I think its basicity is taken $1$, but why?