# 17-1: Common Ion Effect

Hypothes.is Tag = c1403c17 (case sensitive)
Group =
UALRChem1403

## Introduction

The common ion effect is an application of Le Châtelier's Principle to the equilibrium concentration of ionic compounds. We will look at two applications of the common ion effect.

## Application 1: Equilibrium of Acid/Base Buffers

### Type 1: Weak Acid/Salt of Conjugate base

$HA \leftrightharpoons H^+ + A^-$

Le Châtelier's Principle states that if we added the conjugate base of a weak acid to a solution of the weak acid the equilibrium would move to the left to consume that added A-, and the pH would go up as the solution becomes less acidic. This can be done by adding a soluble salt that contains the common ion. Lets start with a solution of 1.0M Hydrofluoric acid and look at the common ion effect by adding 0.10 mole of sodium fluoride to one liter of the solution.

Initially we have:

$$HF$$ $$H^+$$ $$F^-$$
Initial 1.0 0 0
Change -x +x +x
Equilibrium 1.0-x x x

Table $$\PageIndex{1}$$: ICE diagram for pure HF

We can obtain K from table 16.3.1 and since [HF]i >a100K we can ignore the amount HF ionized and $$pH=-log\sqrt{K_{a}[HA]_{i}}$$

$pH=-\log\sqrt{6.3x10^{-4}(1.0)}=1.60$

After adding 0.10mol NaF to one liter

$$HF$$ $$H^+$$ $$F^-$$
Initial 1.0 0 0.10
Change -x +x +x
Equilibrium 1.0-x x 0.10+x

Table $$\PageIndex{2}$$: ICE diagram for pure HF and a salt with its common ion (fluoride)

Noting we can ignore the x in [0.1+x] and 1.0-x] terms because according to Le Châtelier's Principle they will be even smaller because the 0.10 M fluoride pushes the equilibrium to the left.

$K_{a}=\frac{x[0.10+x]}{[1.0-x]}\approx \frac{x(0.10)}{1.0}$

Noting x =H+,and pH=-log H+,equation 17.1.3 can be rewritten in terms of pH.

$pH=-log\left (\frac{ 1.0K_a}{0.1} \right )=-log\left (\frac{ 1.0(6.3x10^{-4})}{0.1} \right )=2.20$

Exercise $$\PageIndex{1}$$

Explain how equations 17.1.2 and 17.1.4 demonstrate Le Châtelier's Principle in terms of the common ion effect.

Eq. 17.1.2 has a pH of 1.60 and represents pure 1M hydrofluoric acid. Eq. 17.1.4 represents that solution after adding 0.10 mole of the common ion fluoride (via sodium fluoride) and that had a pH of 2.20. The solution with the common ion had a higher pH, which means the hydronium ion concentration went down when you added the common ion, which means the reaction consumed the added ion and shifted to the left, as predicted by Châtelier's Principle.

Lets take a closer look at eq. 17.1.4 and the ICE diagram for an acid and its salt. Generically, it would be written as:

$$HA$$ $$H^+$$ $$A^-$$
Initial [HA]i 0 [A-]i
Change -x +x +x
Equilibrium [HA]i-x x [A-]i+x

Table $$\PageIndex{3}$$: Generic ICE diagram for acid and its salt.

$K_{a}=\frac{[H^+]_e[A^-]_e}{[HA]_e}=\frac{x([A^-]_i+x)}{[HA]_i-x}\approx\frac{x[A^-]_i}{[HA]_i}$

What is this saying? It is stating that if the common ion suppresses the ionization to such an extent that you can use the initial concentrations as your equilibrium concentrations. This simplifies the calculations.

### Type 2: Weak Base/Salt of Conjugate Acid

Just as a salt can have the anion of an acid, it can also have the cation that is formed when a weak base reacts with water

$B+H_2O \leftrightharpoons BH^+ + OH^-$

Lets look at the analogous problem for the type 1 acid, but not use the ICE diagram, by considering the weak base ammonia and it's salt, ammonium chloride.

$NH_3(aq)+H_2O(l) \leftrightharpoons NH_4^+(aq) + OH^-(aq)$

We would predict that adding ammonium ion would suppress the ionization of ammonia and thus consume hydroxide, lowering the pH. So what would be the effect of adding 0.10mol of the soluble salt ammonium chloride to a liter of 1.0M ammonia, assuming no change in volume.

Initially, $$pOH=-log\sqrt{K_b[NH_4^+]}$$

$pH=14- \left(-log\sqrt{1.8x10^{-5}[0.1]}\right)=11.11$

After adding Salt and using the logic of eq. 17.1.5:

$K_{b}=\frac{[BH^+]_e[OH^-]_e}{[B]_e}\approx\frac{x[BH^+]_i}{[B]_i} = \frac{[x][NH_4^+]_i}{[NH_3]_i}$

noting x = [OH-] and $$x=\frac{K_b[NH_3]}{[NH_4^+]}$$

$pH = 14-(-log \frac{K_b[NH_3]}{[NH_4^+]})=14+log\frac{1.8x10^{-5}[1.0]}{[0.10]}=10.26$

So the effect is the same as it was with the acid and its salt. That is, adding a salt with a common ion to a base suppresses the ionization of the base, and so the pH of the salt solution is lower than of the base.

In the next section we will see that these solutions function as buffers, as the acid and its conjugate base can coexist, (or the base and its conjugate acid), and this means the solution can have an acid and a base to neutralize any external acid or base that can be added to them. Buffers are important, but before we cover them we will cover the common ion effect on salt solutions.

## Application 2: Solubility of Salts

The effect of adding a soluble salt with a common ion to an insoluble salt of that ion, which causes more of the insoluble salt to precipitate out (ie., suppresses its ionization). OK, this sounds complicated, but it is actually easy to understand. Lets look at barium sulfate, which in chapter 3.4 we learned was an insoluble salt. What does that mean? It means K is a small number and so little barium dissolves, and in fact for barium sulfate, K= 1.1x10-10.

$BaSO_4(s) \rightleftharpoons Ba^{+2}(aq) + SO_4^{-2}(aq) \nonumber \\ K=[Ba^{+2}][SO_4^{-2}]$

Remember this is for a saturated solution, which describes the maximum concentration of dissolved ions that can exist in equilibrium with a solid, and gives a barium concentration of 10 micromolar.

$[Ba^{+2}] = [SO_4^{-2}] = \sqrt{K}=\sqrt{1.1x10^{-10}} =1.0x10^{-5}M$

Le Châtelier's Principle says that if we add a soluble salt of one of the ions we can push the reaction in the direction that consumes that ion, and reduce the concentration of the other ion. So for example, if you want to remove more of a toxic compound like barium from a solution, you could add a soluble sulfate salt, like sodium sulfate.

What is the maximum possible concentration of barium in a solution that is 0.10M Na2SO4 .

BaSO4 $$Ba^{+2}$$ $$SO_4^{-2}$$
Initial Solid 0 0.10M
Change -x +x +x
Equilibrium Solid x 0.10+x

Table $$\PageIndex{2}$$: Note, since barium is a solid it is not part of the equilibrium expression, and K describes the concentration of ions in a saturated solution, so we are writing "solid" to indicate that the concentration of the ions is that which would be in a saturated solution where solid barium sulfate is present. Note for a saturated solution of pure barium sulfate x=1.0x10-5, and when we add sodium sulfate the common ion moves the equilibrium to the left, and the value of x becomes even smaller, and so we can ignore it in the equilibrium expression.

$K=[Ba^{+2}][SO_4^{-2} = x(0.10+x) \approx x(0.10)$

This gives

$Ba^{+2} = x=\frac{K}{0.10}=\frac{1.1x10^{-10}}{0.10} = 1.1x10^{-9}M$

This means you can remove toxic metals like barium from water by adding a soluble salt of their counter ion that makes an insoluble salt with the metal. Barium sulfate is an insoluble salt and sodium sulfate is a soluble salt, and they have the common ion sulfate. Pure barium sulfate is 10 micromolar, and by adding sodium sulfate (a soluble salt) so the sulfate ion concentration is 0.1 M the barium concentration is reduced to 1.1 nanomolar.

Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to: