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$\frac x4$ 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?

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First, the reaction given is not balanced. 3 moles of water should be formed, not 2. However, that will not affect the calculation. Suppose the volume is x ml. The reaction will consume 160 (ml) * 0.25 mole/liter = 40 mmole of HCl. We have x (ml) * 0,1 (mole/liter) = 0,1 * x mmole of salt reacting with HCl. According to the reaction, each mmole of the salt will require 4 mmole of HCl. In order to set up a 1:4 relationship between the salt and the HCl, we have to divide the consumed amount of HCl by 4, i.e. x * 0,1 = 160 * 0,25/4 => x = 100 (ml)

We check:

Salt added: 0,1 (M) * 100 ml = 10 mmole.

HCl added: 0,25 (M) * 160 (ml) = 40 mmole.

Thus, we have a 1:4 relation between the salt and the HCl as the reaction states.

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