Reverse reaction of acid dissociation [closed]








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Consider an acid HA with Ka 10^-10 dissociating into its ions as


HA⇌H+ +A-


My question is-if we introduce solution of pH=3(let's say any weak acid or a buffer) and put ions of A- of 0.1M into it (using strong electrolyte),using expression for Ka(in this case we are operating reverse reaction so k will be reciprocal of Ka) we get concentration of HA to be 10^6 M.Although such a high concentration is theoretically not obtained (assume for my doubt to get clarified) my focus is on the following question (value of 10^6 from FORMULA only):-
How is it CHEMICALLY CORRECT that concentration of products has become more than reactants?While dissociating we can make the argument that some acid is left undissociated so concentration decreases.But how does concentration INCREASE in reverse reaction and what is its chemical significance?
Please help.


(I came across this doubt when I saw a question asking to calculate solubility of AgCN in buffer of pH=3 ksp=1.2×10^-15 and ka of hcn 4.8×10^-10.Solution implied concentration of hcn was much greater than cn from AgCN which confused me as explained above.)




1. The concentration of an acid will never be greater than the concentration of A-. Just because there is not enough of A's to make HA. You are doing something completely wrong in your calcs.
2. In your problem of AgCN, you are not given the amount of AgCN, so how can you actually think of possible concentration of cyanides?
[![enter image description here](https://i.stack.imgur.com/Gr8oP.jpg)](https://i.stack.imgur.com/Gr8oP.jpg)


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