# Calculating Equilibrium Concentrations

$$K_a$$ is an acid dissociation constant, also known as the acid ionization constant. It describes the likelihood of the compounds and the ions to break apart from each other. As we already know, strong acids completely dissociate, whereas weak acids only partially dissociate.  A big $$K_a$$ value will indicate that you are dealing with a very strong acid and that it will completely dissociate into ions. A small $$K_a$$ will indicate that you are working with a weak acid and that it will only partially dissociate into ions.

### General Guide to Solving Problems involving $$K_a$$

Generally, the problem usually gives an initial acid concentration and a $$K_a$$ value.  From there you are expected to know:

1. How to write the $$K_a$$ formula
2. Set up in an ICE table based on the given information
3. Solve for the concentration value, x.
4. Use x to find the equilibrium concentration.
5. Use the concentration to find pH.

### How to write the $$K_a$$ formula

The general formula of an acid dissociating into ions is

$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)} \tag{1}$

with

• $$HA$$ is the acid,
• $$A^-$$ is the conjugate base and
• $$H_3O^+$$ is the the hydronium ion

By definition, the $$K_a$$ formula is written as the products of the equation divided by the reactants of the equation

$K_a = \dfrac{[Products]}{[Reactants]} \tag{2}$

Based off of this general template, we plug in our concentrations from the chemical equation. The concentrations on the right side of the arrow are the products and the concentrations on the left side are the reactants. Using this information, we now can plug the concentrations in to form the $$K_a$$ equation. We then write:

$K_a = \dfrac{[H_3O^+][A^-]}{[HA]} \tag{3}$

Note

The concentration of the hydrogen ion ($$H^+$$) is often used synonymously with the hydrated hydronium ion ($$H_3O^+$$).

• To find a concentration from a pH, we use the formula: $$[H_3O^+]= 10^{-pH}$$
• To find pH from a concentration, we use the formula: $$pH = -\log[H_3O^+]$$
• pH = 7 is neutral
• pH > 7 is basic
• pH < 7 is acidic

Example 1

Calculate the pH of a weak acid solution of 0.2 M HOBr, given:

$HOBr + H_2O \rightleftharpoons H_3O^+ + OBr^-$

$K_a = 2 \times 10^{-9}$

SOLUTION

Step 1: The ICE Table

Since we were given the initial concentration of HOBr in the equation, we can plug in that value into the Initial Concentration box of the ICE chart.  Considering that no initial concentration values were given for H3O+ and OBr-, we can assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes. M stands for molarity.

 HOBr H3O+ OBr- Initial Concentration 0.2 M 0 0 Change in Concentration Equilibrium Concentration

Because we started off without an initial concentration of H3O+ and OBr-, it has to come from somewhere. In the Change in Concentration box, we add a +x because while we do not know what the numerical value of the concentration is at the moment, we do know that it has to be added and not taken away. In contrast, since we did start off with a numerical value of the initial concentration, we know that it has to be taken away to reach equilibrium. Because of this, we add a -x in the HOBr box.

 HOBr H3O+ OBr- Initial Concentration 0.2 M 0 0 Change in Concentration -x M +x M +x M Equilibrium Concentration

Now its time to add it all together! Go from top to bottom and add the Initial concentration boxes to the Change in concentration boxes to get the Equilibrium concentration.

 HOBr H3O+ OBr- Initial Concentration 0.2 M 0 0 Change in Concentration -x M +x M +x M Equilibrium Concentration (0.2 - x) M x M x M

Step 2: Create the $$K_a$$ equation using this equation: $$K_a = \dfrac{[Products]}{[Reactants]}$$

$$K_a = \dfrac{[H_3O^+][OBr-]}{[HOBr-]}$$

Step 3: Plug in the information we found in the ICE table

$$K_a = \dfrac{(x)(x)}{(0.2 - x)}$$

Step 4: Set the new equation equal to the given Ka

$$2 x 10^{-9} = \dfrac{(x)(x)}{(0.2 - x)}$$

Step 5: Solve for x

(x2+ (2 x 10-9x)-(4 x 10-10)

To solve for x, we use the quadratic formula

• $$a=1$$
• $$b=2 \times 10^{-9}$$
• $$c=-4 \times 10^{-10}$$

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \dfrac{-2 \times 10^{-9} \pm \sqrt{(2 \times10^{-9})^2 - 4(1)(-4 \times 10^{-10})}}{2(1)}$

$x=2.0 \times 10^{-5}$

Step 6: Plug x back into the ICE table to find the concentration

$x= [H_3O^+] = 2 \times 10^{-5} \; M$

Step 7: Use the formula using the concentration to find pH

$pH = -\log[H_3O^+] = -\log(2 \times 10^{-5}) = -(-4.69) = 4.69$

$pH= 4.69$

Example 2

For acetic acid, HC2H3O2, the $$K_a$$ value is $$1.8 \times 10^{-5}$$. Calculate the concentration of H3O+ in a 0.3 M solution of HC2H3O2.

SOLUTION

Step 1: The ICE Table

Since we were given the initial concentration of HC2H3O2 in the original equation, we can plug in that value into the Initial Concentration box of the ICE chart. Considering that no initial concentration values were given for $$H_3O^+$$ and $$C_2H_3O_2^-$$, we assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes.

 HC2H3O2 H3O+ C2H3O2 Initial Concentration 0.3 M 0 0 Change in Concentration Equilibrium Concentration

Because we started off without any initial concentration of H3O+ and C2H3O2-, is has to come from somewhere. For the Change in Concentration box, we add a +x because while we do not know what the numerical value of the concentration is at the moment, we do know that it has to be added and not taken away. In contrast, since we did start off with a numerical value of the initial concentration, we know that it has to be taken away to reach equilibrium. Because of this, we add a -x in the $$HC_2H_3O_2$$ box.

 HC2H3O2 H3O+ C2H3O2 Initial Concentration 0.3 M 0 0 Change in Concentration -x M +x M +x M Equilibrium Concentration

Now its time to add it all together! Go from top to bottom and add the Initial concentration boxes to the Change in concentration boxes to get the Equilibrium concentration.

 HC2H3O2 H3O+ C2H3O2 Initial Concentration 0.3 M 0 0 Change in Concentration -x M +x M +x M Equilibrium Concentration (0.3 - x) M x M x M

Step 2: Create the $$K_a$$ equation using this equation:$$K_a = \dfrac{[Products]}{[Reactants]}$$

$$K_a = \dfrac{[H_3O^+][C_2H_3O_2]}{[HC_2H_3O_2]}$$

Step 3: Plug in the information we found in the ICE table

$K_a = \dfrac{(x)(x)}{(0.3 - x)}$

Step 4: Set the new equation equal to the given Ka

$1.8 x 10^{-5} = \dfrac{(x)(x)}{(0.3 - x)}$

Step 5: Solve for x

$(x^2)+ (1.8 \times 10^{-5}x)-(5.4 \times 10^{-6})$

To solve for x, we use the quadratic formula

• $$a=1$$
• $$b=1.8 \times 10^{-5}$$
• $$c=-5.4 \times 10^{-6}$$

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \dfrac{-1.8 \times 10^{-5} \pm \sqrt{(1.8 \times10^{-5})^2 - 4(1)(-5.4 \times 10^{-6})}}{2(1)}$

$x=0.0023$

Step 6: Plug x back into the ICE table to find the concentration

$x= [H_3O^+] = 0.0023\; M$

Example 3

Find the equilibrium concentration of HC7H5O2from a .43 M solution of Benzoic Acid, HC7H5O2

SOLUTION

Given: $$K_a$$ for HC7H5O2= 6.4 x 10-5

Step 1: The ICE Table

 HC7H5O2 H3O+ C7H5O2- Initial Concentration 0.43 M 0 0 Change in Concentration -x +x +x Equilibrium Concentration (0.43-x)M x M x M

Step 2: Create the $$K_a$$ equation using this equation :$$K_a = \dfrac{[Products]}{[Reactants]}$$

$$K_a = \dfrac{[H_3O^+][C_7H_5O_2-]}{[HC_7H_5O_2]}$$

Step 3: Plug in the information we found in the ICE table

$$K_a = \dfrac{(x)(x)}{(0.43 - x)}$$

Step 4: Set the new equation equal to the given Ka

$$6.4 x 10^{-5} = \dfrac{(x)(x)}{(0.43 - x)}$$

Step 5: Solve for x.

x=0.0052

Step 6: Plug x back into the ICE table to find the concentration

[HC7H5O2]= (0.43-x)M

[HC7H5O2]= (0.43-0.0052)M

Example 4

For a 0.2M solution of Hypochlorous acid, calculate all equilibrium concentrations.

SOLUTION

Given: $$K_a = 3.5 \times 10^{-8}$$

Step 1: The ICE Table

 HOCl H3O+ OCl- Initial Concentration 0.2 M 0 0 Change in Concentration -x M +x M +x M Equilibrium Concentration (0.2 - x) M x M x M

Step 2: Create the $$K_a$$ equation using this equation: $$K_a = \dfrac{[Products]}{[Reactants]}$$

$$K_a = \dfrac{[H_3O^+][OCl-]}{[HOCl-]}$$

Step 3: Plug in the information we found in the ICE table

$$K_a = \dfrac{(x)(x)}{(0.2 - x)}$$

Step 4: Set the new equation equal to the given Ka

$$3.5 x 10^{-8} = \dfrac{(x)(x)}{(0.2 - x)}$$

Step 5: Solve for x

x=8.4 x 10-5

Step 6: Plug x back into the ICE table to find the concentration

[HOCl]= [(.2)-(8.4 x 10-5)]=.199

[H3O+]=8.4 x 10-5

[OCl-]=8.4 x 10-5

Example 5

Calculate the pH from the equilibrium concentrations of [H3O+] in Example 4.

SOLUTION

Given:

[HOCl]=0.199

[H3O+]=8.4 x 10-5

[OCl-]=8.4 x 10-5

Step 1: Use the formula using the concentration of [H3O+] to find pH

$pH = -\log[H3O+] = -\log(8.4 x 10^{-5}) = 4.08$