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.)

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    $\begingroup$ Sure enough, $10^6\rm M$ screams nonsense, no matter how it came about. $\endgroup$ Oct 9, 2017 at 7:02
  • $\begingroup$ Can you please write down what exactly you want to add to what? I recommend you also translate that into specific amounts of real chemicals. It will become very clear where you are going wrong then. $\endgroup$
    – Karl
    Oct 9, 2017 at 7:20

1 Answer 1

  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
  • $\begingroup$ Can you explain that last sentence "it's still a buffer" $\endgroup$ Oct 9, 2017 at 8:01
  • $\begingroup$ That means the concentration of H+ wouldn't change no matter what happens, as long as we assume it's a good buffer solution, and no "external" equilibria influence it. $\endgroup$ Oct 9, 2017 at 8:06

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