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If I have $0.01~\mathrm{M}$ $\ce{HBr}$ in $0.09~\mathrm{M}$ $\ce{KBr}$, why would the presence of $\ce{Br-}$ ions from the $\ce{KBr}$ not reduce the dissociation of $\ce{HBr}$ based on Le Chatelier's principle? Or would it?

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The $\ce{Br-}$ would make less $\ce{HBr}$ dissociate, But $\ce{HBr}$ is such a strong acid ($\mathrm{pKa}$ is $\mathrm{-9}$) that still only a trivial amount would be molecular.

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As pointed out be DavePhD, in reality there will be no real detectable change in the dissociation of $\ce{HBr}$ since it is such a strong acid. I will attempt to give a more theoretical answer to this question using equilibrium constants.

We have the following chemical equations (note that we assume $\ce{KBr}$ completely disassociates) : $$\ce{HBr \rightleftharpoons H+ + Br-}$$ $$\ce{KBr -> K+ + Br-}$$

We have two equations: $$\ce{K_{a}\ = \ \frac{[H^+][Br^{-}]}{[HBr]}}$$ $$\ce{[Br^{-}] = [Br^{-}_{HBr}] + [Br^{-}_{KBr}]}$$ The extent of disassocation of $\ce{HBr}$ is given by value of $\ce{[Br^{-}_{HBr}]}$.

By substituting the second equation into the first equation and rearranging: $$\ce{K_{a}\ = \ \frac{[H^+]([Br^{-}_{HBr}] + [Br^{-}_{KBr}])}{[HBr]}}$$ $$\ce{[H^+][Br^{-}_{HBr}]\ =\ K_{a}[HBr] - [H^+][Br^{-}_{KBr}]}$$ $$\ce{[Br^{-}_{HBr}]\ =\ \frac{K_{a}[HBr] - [H^+][Br^{-}_{KBr}]}{[H^+]}}$$ From this equation, it can clearly be seen that the disassociation of $\ce{HBr}$ decreases by adding $\ce{KBr}$. However since the Ka value is ${10^9}$, the second term in the numerator is negligible compared to the first term and the equation can be reduced to: $$\ce{[Br^{-}_{HBr}]\ \approx \ \frac{K_{a}[HBr] }{[H^+]}}$$ As you can see, this expression is independent of the concentration of $\ce{KBr}$ which illustrates DavePhD's point that it really won't effect the disassociation of $\ce{HBr}$.

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