Let $\ce{HA}$ ($x$ molar) be a strong electrolyte and $\ce{HB}$ ($c$ molar) — weak electrolyte. As $\ce{HA}$ is strong electrolyte, it dissociates completely:

$$ \begin{array}{lcc} \ce{&HA &-> &H+ &+ &A-} \\ \text{initially} & x && 0 && 0 \\ \text{after} & 0 && x && x \end{array} $$

As HB is weak electrolyte

$$ \begin{array}{lcc} \ce{&HB &-> &H+ &+ &B-} \\ \text{initially} & c && 0 && 0 \\ \text{equilibrium} & c - cα && cα && cα \end{array} $$


$$K_\mathrm{a} =\frac{cα\cdot (cα + x)}{c - cα}$$

Now my doubt is how can we add the concentration of $\ce{H+}$ together as their volume can be different. Suppose $x = \pu{1 M}$, and $cα = \pu{0.01 M},$ then how can be the resultant concentration be $\pu{1.01 M}?$ Can someone explain in detail?

Edit: $\alpha$ is the degree of dissociation Now let us suppose we had a 'x' molar strong electrolyte and a 'c' having a common ion let us suppose H. Now while writing Ka for the dissocition of weak electrolyte as given in equation $$ \begin{array}{lcc} \ce{&HB &-> &H+ &+ &B-} \\ \text{initially} & c && 0 && 0 \\ \text{equilibrium} & c - cα && cα && cα \end{array} $$

Then $$K_\mathrm{a} =\frac{[B] \cdot [H] }{[HB] }$$ Now due to common ion effect we have to consider the total concentration of H we get which gives us

$$K_\mathrm{a} =\frac{cα\cdot (cα + x)}{c - cα}$$ Now how did the concentration of H become $$(cα + x)$$ Let us now suppose instead of x we had a 2 molar strong electrolyte (HA) so the concentration of H that comes from HA is 2 molar. Now for the weak electrolyte let us suppose we get 0.01M as the concentration of H$(cα) $ that comes from HB now according to the equation $$K_\mathrm{a} =\frac{cα\cdot (cα + x)}{c - cα}$$ We get the total concentration of H is (cα + x). Using our supposed value we get total concentration of H as 2+0.01=2.01M but how can we just add the two concentration. Just like in if we had a 1M solution1 and a 2M solution2 we would not get the total concentration of the two solutions after mixing as 3M

  • $\begingroup$ Are x and c the concentrations before mixing or after? What are the different volumes you are talking about? $\endgroup$ – Karsten Theis May 16 '19 at 4:38
  • $\begingroup$ Initial concentration $\endgroup$ – Satwik May 16 '19 at 5:15
  • $\begingroup$ It is a common ion, so you should have a single variable for the $\ce{H+}$ concentration, and it should appear in the equilibrium expression. The protonation state of the weak electrolyte (just like a pH indicator) will be influenced by the pH of the solution. Another consequence is that the concentration of $\ce{H+}$ will be different from the concentration of $\ce{B-}$. $\endgroup$ – Karsten Theis May 16 '19 at 5:23
  • $\begingroup$ @Satwik It's not the concentration that we add, we add the moles. If the volume is 1 L, then naturally concentrations get added. $\endgroup$ – Light Yagami Dec 14 '20 at 7:50

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