Common ion effect

Dissolve sodium acetate into a solution of acetic acid and something odd happens: the acid barely ionises anymore. Nothing about the acid molecule changed, and KaK_a is still 1.8×10−51.8\times10^{-5} — yet [H+][H^+] collapses to a tiny fraction of what it was. The culprit is the acetate ion, CH3COO−\text{CH}_3\text{COO}^-, which the salt dumps into solution in bulk. Because that same ion also appears on the product side of the acid's own equilibrium, flooding it in from outside pushes the equilibrium backward, and ionisation is suppressed. This is the common-ion effect — Le Chatelier's principle applied to ionic equilibrium, and one of the most exam-tested ideas in equilibrium chemistry because it is also the mechanism behind buffers and behind the lowered solubility of sparingly soluble salts. This article builds the ICE-table shortcut that lets you calculate [H+][H^+] or [OH−][OH^-] in a common-ion mixture in under a minute, works through the numbers, and flags the traps examiners love to set.

The core insight

Adding a strong electrolyte that supplies an ion already present in a weak electrolyte's equilibrium shifts that equilibrium backward (Le Chatelier), suppressing ionisation without changing KaK_a or KbK_b — quantitatively, [H+]≈Ka ca/cs[H^+] \approx K_a\,c_a/c_s.

The core idea

1. It is Le Chatelier's principle, nothing more exotic. A weak acid sits in dynamic equilibrium:

HA⇌H++A−HA \rightleftharpoons H^+ + A^-

If you add a strong electrolyte such as the salt NaANaA, it dissociates completely and dumps a large, sudden extra supply of A−A^- into the mixture. The system reacts to relieve that stress by consuming some of the extra A−A^- (and some H+H^+), driving the equilibrium backward, toward HAHA.

2. The ion must already belong to the equilibrium. This is what makes it a common ion — it is not any old ion, but specifically one that appears in the weak electrolyte's own dissociation equation. Adding NaClNaCl to acetic acid does nothing of this kind, because Cl−Cl^- plays no part in the acetic-acid equilibrium.

3. The observable result is a drop in α\alpha and in [H+][H^+] (or [OH−][OH^-]). The degree of ionisation α\alpha of the weak electrolyte falls, and so does the concentration of the ion it was producing — even though the total amount of weak electrolyte present is unchanged. For a weak base BOH⇌B++OH−BOH \rightleftharpoons B^+ + OH^-, adding a common ion B+B^+ (from a strong salt like BClBCl) suppresses ionisation of BOHBOH and lowers [OH−][OH^-] in exactly the same way.

4. KaK_a itself never changes. Only the position of equilibrium shifts. This distinction — between the constant KaK_a and the variable quantities [H+][H^+], [A−][A^-], α\alpha — is where most marks are lost in exams.

Figure
Two beakers compared, plain acetic acid ionising more than acetic acid with sodium acetate added, showing the equilibrium arrow pushed back toward the un-ionised acid
Common ion pushes the ionisation equilibrium backward plain 0.10 M acetic acid HA ⇌ H⁺ + A⁻ alpha approx 1.34 percent plus 0.10 M sodium acetate HA ⇌ H⁺ + A⁻ alpha approx 0.018 percent purple = extra A minus from the salt, dominating the mixture
Derivation ACommon-ion suppression of H plus

Goal: for a weak acid HAHA of initial concentration cac_a mixed with a strong electrolyte that supplies an initial common-ion concentration csc_s of A−A^-, find an approximate expression for [H+][H^+] at equilibrium in terms of KaK_a, cac_a, and csc_s, and show it is far smaller than the ionisation of HAHA alone.

Step 1. Write the equilibrium and set up the ICE table. Before any change, [HA]0=ca[HA]_0 = c_a and [A−]0=cs[A^-]_0 = c_s (the salt, a strong electrolyte, is taken to dissociate completely, so it contributes its full stated concentration to A−A^- immediately). Let xx be the amount of HAHA that ionises. Then at equilibrium [HA]=ca−x[HA] = c_a - x, [H+]=x[H^+] = x, and [A−]=cs+x[A^-] = c_s + x.

Step 2. Substitute into the equilibrium expression Ka=[H+][A−][HA]K_a = \dfrac{[H^+][A^-]}{[HA]}:

Ka=x(cs+x)ca−xK_a = \dfrac{x(c_s + x)}{c_a - x}

Step 3. Apply the key approximation that makes this tractable. Because the common ion strongly suppresses further ionisation, xx is very small compared to both cac_a and csc_s, so ca−x≈cac_a - x \approx c_a and cs+x≈csc_s + x \approx c_s:

Ka≈x cscaK_a \approx \dfrac{x\, c_s}{c_a}

Step 4. Solve for x=[H+]x = [H^+]:

[H+]=x≈Kacacs(1)[H^+] = x \approx K_a\dfrac{c_a}{c_s} \qquad (1)

where: KaK_a is the acid dissociation constant, cac_a is the initial (stated) concentration of the weak acid, and csc_s is the initial concentration of the common ion supplied by the strong electrolyte.

Step 5. Compare with the weak acid alone (no common ion), where [H+]≈Kaca[H^+] \approx \sqrt{K_a c_a}. Because Equation (1) divides by csc_s instead of taking a square root, adding even a modest common-ion concentration drives [H+][H^+] down sharply — this is the algebraic signature of the common-ion effect, and it is the same rearrangement that later gives the Henderson–Hasselbalch equation for buffers. The identical algebra, with KbK_b, cbc_b (base) and csc_s (common cation), gives [OH−]≈Kb cb/cs[OH^-] \approx K_b\,c_b/c_s for a weak base with a common cation.

Worked example

Suppression of [H+][H^+] and degree of ionisation by a common ion. A solution is 0.10 M0.10\ M in acetic acid (Ka=1.8×10−5K_a = 1.8\times10^{-5}) and also 0.10 M0.10\ M in sodium acetate. Find [H+][H^+] and α\alpha for the acetic acid, and compare with pure 0.10 M0.10\ M acetic acid alone.

Step 1. Sodium acetate dissociates completely: CH3COONa→CH3COO−+Na+\text{CH}_3\text{COONa} \rightarrow \text{CH}_3\text{COO}^- + Na^+, so cs=0.10 Mc_s = 0.10\ M before any further ionisation of the acid.

Step 2. Set up the ICE table for CH3COOH⇌H++CH3COO−\text{CH}_3\text{COOH} \rightleftharpoons H^+ + \text{CH}_3\text{COO}^- with ca=0.10 Mc_a = 0.10\ M, cs=0.10 Mc_s = 0.10\ M: equilibrium values [CH3COOH]≈0.10[\text{CH}_3\text{COOH}] \approx 0.10, [H+]=x[H^+] = x, [CH3COO−]≈0.10[\text{CH}_3\text{COO}^-] \approx 0.10.

Step 3. Apply Equation (1): [H+]≈Kacacs=1.8×10−5×0.100.10=1.8×10−5 M[H^+] \approx K_a\dfrac{c_a}{c_s} = 1.8\times10^{-5}\times\dfrac{0.10}{0.10} = 1.8\times10^{-5}\ M.

Step 4. Degree of ionisation α=x/ca=1.8×10−4\alpha = x/c_a = 1.8\times10^{-4}, i.e. 0.018%.

Step 5. Compare: pure 0.10 M0.10\ M acetic acid alone gives [H+]≈1.34×10−3 M[H^+] \approx 1.34\times10^{-3}\ M and α≈1.34%\alpha \approx 1.34\% (from [H+]≈Kaca[H^+]\approx\sqrt{K_ac_a}). With the common ion present, both are about 75 times smaller — the dramatic suppression Le Chatelier's principle predicts. Final answer: [H+]≈1.8×10−5 M[H^+] \approx 1.8\times10^{-5}\ M, α≈0.018%\alpha \approx 0.018\%.

Competitive stretch

Change only the base concentration: find [OH−][OH^-] for a solution 0.20 M0.20\ M in NH4OH\text{NH}_4\text{OH} (Kb=1.8×10−5K_b = 1.8\times10^{-5}) and 0.10 M0.10\ M in NH4Cl\text{NH}_4\text{Cl}. Using the base analogue of Equation (1), [OH−]≈Kb cb/cs=(1.8×10−5)(0.20)/(0.10)=3.6×10−5 M[OH^-] \approx K_b\,c_b/c_s = (1.8\times10^{-5})(0.20)/(0.10) = 3.6\times10^{-5}\ M — again far below the ≈1.9×10−3 M\approx 1.9\times10^{-3}\ M that pure 0.20 M0.20\ M NH4OH\text{NH}_4\text{OH} alone would give.

Common traps

  • Reducing the salt's contribution by xx. The common-ion source is a strong electrolyte, assumed fully dissociated — its contribution enters the ICE table as its complete stated concentration, never as (cs−x)(c_s - x).
  • Skipping the negligibility check. The approximations ca−x≈cac_a - x \approx c_a and cs+x≈csc_s + x \approx c_s are valid because the common ion suppresses ionisation strongly; state briefly why x≪ca,csx \ll c_a, c_s rather than assuming it silently.
  • Thinking KaK_a or KbK_b changes. It does not — only [H+][H^+], [A−][A^-], and α\alpha shift; the equilibrium constant is a function of temperature alone.
  • Applying the effect to strong-electrolyte ionisation. Strong acids/bases/salts are already essentially completely ionised, so there is no meaningful equilibrium left to push back; the common-ion effect only applies to a weak electrolyte's own equilibrium.

Ask the tutor: Want to see this exact algebra turn into the Henderson–Hasselbalch equation for buffer pH, or into a solubility-product suppression problem?

Does adding a common ion change the value of Ka?

No. KaK_a (like KbK_b) depends only on temperature. Adding a common ion changes the concentrations of the species at equilibrium — it shifts where the equilibrium sits — but the ratio [H+][A−]/[HA][H^+][A^-]/[HA] still evaluates to the same KaK_a once the system re-settles. Students often lose marks by writing that KaK_a "decreases due to suppression"; it is [H+][H^+] and α\alpha that decrease, not KaK_a.

Why is the approximation ca minus x approximately ca considered safe here, but not always for a weak acid alone?

For a weak acid alone, the approximation ca−x≈cac_a - x \approx c_a is safe only when α\alpha is small (roughly under 5%), and you are expected to check it. With a common ion present, xx is suppressed to be even smaller than it would be alone (here about 75 times smaller), so the same approximation becomes even more reliable — but it is still good practice to state that x≪cax \ll c_a and x≪csx \ll c_s rather than assume it without comment.

How does this connect to buffers and to solubility product later in the syllabus?

A buffer is essentially a common-ion mixture examined from a different angle: the same ICE-table algebra that gives [H+]≈Kaca/cs[H^+] \approx K_a c_a/c_s rearranges (taking logarithms) into the Henderson–Hasselbalch equation used to calculate buffer pH. The common-ion effect also explains why adding a salt like NaClNaCl lowers the solubility of AgClAgCl in a saturated solution — the extra Cl−Cl^- pushes the dissolution equilibrium backward exactly as here, and that quantitative link is developed fully under solubility product.

Before this

Where this leads