Calculate the percent ionization of the in a 0.10 m solution?

Question: Calculate the percent ionization of the in a 0.10 m solution?

How to Calculate the Percent Ionization of a Weak Acid

The percent ionization of a weak acid is a measure of how much it dissociates into ions when dissolved in water. The higher the percent ionization, the stronger the acid. To calculate the percent ionization of a weak acid, we need to know its initial concentration and its equilibrium concentration of hydronium ions.

The formula for percent ionization is:

Percent ionization = ([H3O+]eq / [HA]0) x 100%

where [H3O+]eq is the equilibrium concentration of hydronium ions and [HA]0 is the initial concentration of the weak acid.

To find [H3O+]eq, we can use the acid dissociation constant (Ka) of the weak acid and an ICE table. The Ka is a measure of how much the acid ionizes in water and is given by:

Ka = ([H3O+]eq x [A-]eq) / [HA]eq

where [A-]eq is the equilibrium concentration of the conjugate base of the acid.

An ICE table shows the initial, change, and equilibrium concentrations of the reactants and products in a chemical reaction. For a weak acid HA, the ionization reaction is:

HA + H2O ⇌ H3O+ + A-

The ICE table for this reaction is:

| HA   | H2O  | H3O+ | A-  |

|------|------|------|-----|

| I    | -    | 0    | 0   |

| C    | -    | +x   | +x  |

| E    | -    | x    | x   |

where x is the amount of HA that ionizes.

To solve for x, we can substitute the equilibrium concentrations into the Ka expression and solve for x using algebra or a quadratic formula. Once we find x, we can plug it into the percent ionization formula and get our answer.

Let's see an example of how to apply this method.

Example: Calculate the percent ionization of a 0.10 M solution of acetic acid (CH3COOH). The Ka of acetic acid is 1.8 x 10^-5.

Solution:

Step 1: Write the balanced ionization reaction and the Ka expression.

CH3COOH + H2O ⇌ H3O+ + CH3COO-

Ka = ([H3O+]eq x [CH3COO-]eq) / [CH3COOH]eq

Step 2: Set up an ICE table with the given initial concentration and assume x for the change and equilibrium concentrations.

| CH3COOH | H2O     | H3O+    | CH3COO- |

|---------|---------|---------|---------|

| I       | 0.10 M  | -       | 0 M     |

| C       | -x      | -       | +x      |

| E       | 0.10-x  | -       | x       |

Step 3: Substitute the equilibrium concentrations into the Ka expression and solve for x.

Ka = ([H3O+]eq x [CH3COO-]eq) / [CH3COOH]eq

1.8 x 10^-5 = (x x x) / (0.10-x)

Simplifying and rearranging, we get:

1.8 x 10^-5 (0.10-x) = x^2

Expanding and moving everything to one side, we get:

x^2 + 1.8 x 10^-5 x - 1.8 x 10^-6 = 0

This is a quadratic equation that we can solve using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

where a = 1, b = 1.8 x 10^-5, and c = -1.8 x 10^-6.

Plugging in these values, we get:

x = (-1.8 x 10^-5 ± √((1.8 x 10^-5)^2 - 4(1)(-1.8 x 10^-6))) / (2(1))

x = (-1.8 x 10^-5 ± √(3.24 x 10^-10 + 7.2 x 10^-6)) / 2

x = (-1.8 x 10^-5 ± √(7.20324 x 10^-6)) / 2

x = (-1.8 x 10^-5 ± 0.002683) / 2

We get two possible values for x, but we only take the positive one since the negative one does not make sense for a concentration.

x = (-1.8 x 10^-5 + 0.002683) / 2

x = 0.001341 M

This is the equilibrium concentration of H3O+ and CH3COO-.

Step 4: Plug x into the percent ionization formula and get the answer.

Percent ionization = ([H3O+]eq / [HA]0) x 100%

Percent ionization = (0.001341 / 0.10) x 100%

Percent ionization = 1.341%

The percent ionization of a 0.10 M solution of acetic acid is 1.341%.