# Common ion effect

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

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

Then

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

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

• Are x and c the concentrations before mixing or after? What are the different volumes you are talking about? – Karsten Theis May 16 '19 at 4:38
• Initial concentration – Satwik May 16 '19 at 5:15
• 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 $\ce{B-}$. – Karsten Theis May 16 '19 at 5:23
• @Satwik It's not the concentration that we add, we add the moles. If the volume is 1 L, then naturally concentrations get added. – Light Yagami Dec 14 '20 at 7:50