It is to be noted that the strong acids and bases dissociate or ionize completely so their percent ionization is 100%. This means that each hydrogen ions from each acidic molecule or the hydroxide ions from each basic molecule separate or donate. So, only the percent ionization of weak acid or base is calculated. The calculation of the percent ionization is one way of quantifying how far a weak acid has dissociated from the solution. When considering solutions it becomes much more reasonable to learn of the ionized percentage against the concentrations or the equilibrium constant.

As weak acid and base solutions are partially ionized in water which lead to too many charged and uncharged species in dynamic stabilization. The ionized percentage is the proportion of the initial ionisation compound. So, the concentration of the ion in solution is compared with the initial neutral species concentration. Weak acids like hydrofluoric acid, acetic acid, formic acid, water, etc. and weak bases like ammonia, methyl amine, etc. ionize in water in small quantities. It is very simple to compute the percent ionization of acid or base which can allow you to know the behaviour of certain weak acids and bases.

This article will show you the percent ionization formula, how to calculate it step by step and give you examples on how the formula can be utilized.

**Percent Ionization Formula**

Percent ionization can be computed by dividing the concentration of ionized acid or base in equilibrium by the original concentration of the solution times 100 percent. The concentration in equilibrium can be determined through the equilibrium constant inherent to the solution itself and upon its complete ionization. The formula is given by;

For acid,

Percent ionization=H3O+eqHA0100 %

For base,

Percent ionization=OH-eqHA0100 %

**How to calculate Percent Ionization**

The steps to calculate percent abundance are given below;

- Firstly, write the balanced acid or base dissociation/ionization reaction
- Secondly, set down the expression for dissociation constant Ka for acid or Kb for base.
- Compute [H3O+] and conjugate base for acid or OH- and conjugate acid for base at equilibrium
- Finally, calculate percent ionization

**Ionization of acids**

A strong acid ionizes in water entirely while a weak acid only partly ionizes. The equilibrium Constant for ionization of an acid defines its Acid Ionisation Constant (K_{a}). However, the stronger the acid, the greater will be the acid ionisation constant (K_{a}) meaning that a strong acid is a better proton donor.

*Figure 1.** Strong acids completely dissociate in water.*

The weak acid solution of water is the nonionized acid, hydronium ion and the acid conjugate base with the highest concentration of nonionized acid. This raises the level of hydronium ion in an aqueous solution of weak acid.

**Example 1. Calculating Percent Ionization of a Weak Acid**

If acetic acid (CH3COOH) has a Ka of 1.8*10-5 at 250C, what is the percent ionization of acetic acid in a 1.00 M solution?

Let’s go through this example step-by-step.

**Step 1: Write the balanced acid dissociation/ionization reaction**

Let’s firstly write the balanced ionization reaction of CH3COOH in water. The acetic acid can donate a proton to water to form CH3COO-(aq).

CH3COOHaq+H2OlH3O+aq+CH3COO-(aq)

**Step 2: Then, set down the expression for dissociation constant **Ka** for acid**

As we have the equation from step 1, now can write the dissociation constant Ka expression of acetic acid:

Ka=H3O+[CH3COO-][CH3COOH]=1.8*10-5

**Step 3: Compute **[H3O+]** and conjugate base for acid **[CH3COO-]** at equilibrium**

As we know acetic acid is weak acid so the complete ionization of weak acid is not detected. Now, we can use an ICE (Initial, Change, Equilibrium) table to evaluate algebraic expressions for the equilibrium concentrations in Ka expression.

CH3COOHaq + H2Ol ↔ H3O+aq + CH3COO-aq

I(M) | 1.00 | 0 | 0 | 0 |

C(M) | -x | 0 | +x | +x |

E(M) | 1.00-x | 0 | x | x |

Now, putting the equilibrium concentrations into our Ka expression, we have;

Ka=(x)(x)(1.00-x)=1.8*10-5

Next, we can simplify the expression by following way;

x21.00-x=1.8*10-5

This seems to be a quadratic equation so it can be solved by using a quadratic formula or approximation method.

Since the right side of Ka of acetic acid is very small i.e. 10-5 which is less than 10-4 , we can use the approximation method and ignore x on the denominator on the left side. Then, we get,

x21.00=1.8*10-5

or x=1.8*10-5×1.00=4.2*10-3 M

Therefore, H3O+aq=CH3COO-aq=4.2*10-3 M

**Step 4: Finally, calculate the percent ionization**

Now, to find the percent ionization, let us use the equilibrium expressions we have get in Step 3.

Percent ionization=H3O+eqHA0100 %

or Percent ionization=H3O+eqCH3COOHinitial100 %=4.2*10-31.00100 %=0.42 %

Hence, 0.42 % of the acetic acid (CH3COOH) in solution has ionized into H+ & CH3COO- ions.

**Ionization of bases**

There are strong bases such as sodium hydroxide, lithium hydroxide etc. that get fully dissociated into their ions in an aqueous solution. The ionization-base constant i.e. K_{b} relates to the equilibrium constant for the ionization of a base. So we can tell that a strong basis indicates a good acceptor of protons

In equilibrium, a solution of a weak base in water is a combination of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the highest concentration. Therefore, a weak base raises the hydroxide ion concentration in an aqueous solution.

**Example 2. Calculating Percent Ionization of a Weak Base**

What is the percent ionization of a 1.5 M solution of ammonia? (Kb= 1.8*10-5).

The step by step solution of the problem is given below:

**Step 1: Write the balanced base ionization reaction**

Let’s firstly write the balanced base ionization reaction for ammonia. The ammonia can accept a proton from water to form ammonium, NH4+(aq).

NH3aq+H2OlNH4+aq+OH-(aq)

**Step 2: Then, set down the expression for dissociation constant **Kb** for base**

As we have the equation from step 1, now can write the dissociation constant Kb expression of ammonia:

Kb=NH4+[OH-][NH3]=1.8*10-5

**Step 3: Compute **NH4+ & [OH-]** at equilibrium**

As we know ammonia is weak base so the complete ionization of weak base is not detected. Now, we can use an ICE (Initial, Change, Equilibrium) table to evaluate algebraic expressions for the equilibrium concentrations in Kb expression.

NH3aq + H2Ol NH4+aq + OH-(aq)

I(M) | 1.50 | 0 | 0 | 0 |

C(M) | -x | 0 | +x | +x |

E(M) | 1.50-x | 0 | x | x |

Now, putting the equilibrium concentrations into our Kb expression, we have;

Kb=(x)(x)(1.50-x)=1.8*10-5

Next, we can simplify the expression by following way;

x21.50-x=1.8*10-5

This seems to be a quadratic equation so it can be solved by using a quadratic formula or approximation method.

Since the right side of Kb of ammonia is very small i.e. 10-5 which is less than 10-4 , we can use the approximation method and ignore x on the denominator on the left side. Then, we get,

x21.50=1.8*10-5

or x=1.8*10-5×1.50=5.2*10-3 M

Therefore, NH4+aq=OH-aq=5.2*10-3 M

**Step 4: Finally, calculate the percent ionization**

Now, to find the percent ionization, let us use the equilibrium expressions we have got in Step 3.

Percent ionization=OH-eqHA0100 %

or Percent ionization=OH-eqNH3initial100 %=5.2*10-31.50100 %=0.34 %

Hence, 0.34 % of the ammonia (NH3) in solution has ionized into NH4+ & OH- ions.

**Example 3. Calculating Percent Ionization from pH**

Calculate the percent ionization of a 0.125 M solution of nitrous acid (weak acid) with a pH of 2.09.

Here, firstly write the balanced chemical equation of ionization reaction of HNO2 in water.

HNO2aq+H2OlH3O+aq+NO2-(aq)

Then, we have given pH = 2.09

As **pH** is a measure of hydrogen ion concentration, a measure of the acidity or alkalinity of a solution so we have,

pH=-log(H3O+)

or 2.09=-log H3O+

or 10-2.09=H3O+

or H3O+=8.1*10-3 M

Finally, percent ionization of nitrous acid is given by;

Percent ionization=8.1*10-30.125100%=6.5%