General Guide to Solving Problems involving Ka

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 Ka 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 Ka formula  

The general formula of an acid dissociating into ions is

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

(HA=an acid, A^-= conjugate base, H₃O⁺= hydronium ion)

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

\(K_a = \dfrac{[Products]}{[Reactants]}\)

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 Ka equation.

We then write:

 \(K_a = \dfrac{[H_3O^+][A^-]}{[HA]}\)

note: The concentration of the Hydrogen ion [H+] is used synonymously with the concentration of the Hydronium ion [H30+] 

Formulas

• To find a concentration from a pH, we use the formula: [H3O+]= 10^(-pH)
• To find pH from a concentration, we use the formula: pH = -log[H3O+]
  • 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 x 10^-9

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 H₃0⁺ 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. 

Because we started off without an initial concentration of H₃0⁺ 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.

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. 

Step 2: Create the Ka 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)

In order to solve for x, we use the quadratic formula

a=1, b=2 x 10^-9, c=-4 x 10^-10

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

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

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

Answer: pH= 4.69

Example #2

 For acetic acid, HC₂H₃O₂, the \(K_a\) value is 1.8 x 10^-5. Calculate the concentration of H₃O⁺ in a .3 M solution of HC₂H₃O_2.

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 H30+ and C₂H₃O_2^-, we assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes.  

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 HC2H3O2 box.

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. 

Step 2: Create the Ka 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 K_a

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

Step 5: Solve for x

(x²)+ (1.8 x 10^-5x)-(5.4 x 10^-6)

In order to solve for x, we use the quadratic formula

a=1, b=1.8x 10^-5, c=-5.4x10^-6

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

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

\(x= [H_3O^+] = .0023 M\)

Now try some on your own.  

Example #3

• The Determination of the pKa of Multiprotic, Weak Acids
• Determination of the Equilibrium Constants of a Weak Acid: An Experiment for analytical or Physical Chemistry
• The pKa of a Weak Acid as a Function of Temperature and Ionic Strength
• Graphical Correlation between pH Values, Molarities, and Dissociation Constants of Weak Acids
• Hawkes, Stephen J. "Easy Derivation of pH (pK_a1 + pK_a2) / 2 Using Autoprotolysis of HA^-: Doubtful Value of the Supposedly More Rigorous Equation." J. Chem. Educ. 2000 77 1183.
• van Lubeck, Henk. "Why Not Replace pH and pOH by Just One Real Acidity Grade, AG?." J. Chem. Educ. 1999 76 892.

Contributors

• Kellie Berman (UCD), Alysia Kreitem (UCD)

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