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Ratio of [HA^(+)] in 1L of 0.1M H(3)A so...

Ratio of `[HA^(+)]` in `1L` of `0.1M H_(3)A` solution `[K_(a_(1))=10^(-5),K_(a_(2))=10^(-8)K_(a_(3))=10^(-11)]`& upon addition of `0.1`mole `HCl` to it will be :

A

`10`

B

`100`

C

`1000`

D

`10,000`

Text Solution

AI Generated Solution

To solve the problem, we need to find the ratio of \([HA^+]\) in a \(1L\) solution of \(0.1M H_3A\) before and after the addition of \(0.1\) moles of \(HCl\). Given the dissociation constants \(K_{a1} = 10^{-5}\), \(K_{a2} = 10^{-8}\), and \(K_{a3} = 10^{-11}\), we can approach the problem step by step. ### Step 1: Initial Concentration of Species In a \(0.1M\) solution of \(H_3A\), we can assume that initially, the concentration of \(H_3A\) is \(0.1M\). The dissociation of \(H_3A\) can be represented as follows: 1. \(H_3A \rightleftharpoons H^+ + H_2A^-\) with \(K_{a1} = 10^{-5}\) 2. \(H_2A^- \rightleftharpoons H^+ + HA^{2-}\) with \(K_{a2} = 10^{-8}\) ...
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