Scientific Tutor

How does Equilibrium relate to acids and bases?  Ka and Kb

Ka and Kb are the representations of the equilibrium constant for acid or base reactions. Again, there is nothing different in the math or equation set up for Ka or Kb. All equilibrium acid or based chemical equations are set up in one direction like the examples below; where the acid or base together is shown on the reactants side and the ions of the acid or base is shown on the products side. This will NEVER change for Ka or Kb equilibrium equations.

 

Acid:

HCl_(aq) <—> H⁺_(aq) + Cl^–_(aq)

Ka =	[ H+ ] [ Cl– ]
	[ HCl ]

 

Base:

NaOH_(aq) <—> Na⁺_(aq) + OH^–_(aq)

Kb =	[ Na+ ] [ OH– ]
	[ NaOH ]

 

The generic equations for Ka and Kb look like the ones below. They are useful for trying to form equations when the you get complicated acid and base questions: HA represents the acid, A- represents the negative ion that the acid produces, BOH represents the base, B+ represents the positive ion that the base produces.

 

HA_(aq) <—> H⁺_(aq) + A^–_(aq)

Ka =	[ H+ ] [ A– ]
	[ HA ]

 

BOH_(aq) <—> B⁺_(aq) + OH^–_(aq)

Kb =	[ B+ ] [ OH– ]
	[ BOH ]

 

Most acid or base equilibria are pretty straightforward to follow and match up the generic formulas.

 

Examples: Create the equilibrium equation from the chemical equations below.

 

HF_(aq) <—> H⁺_(aq) + BF^–_(aq)

Ka =	[ H+ ] [ F– ]
	[ HF ]

 

KOH_(aq) <—> K⁺_(aq) + OH^–_(aq)

Kb =	[ K+ ] [ OH– ]
	[ KOH ]

 

Ba(OH)_2(aq) <—> Ba²⁺_(aq) + 2 OH^–_(aq)

Kb =	[ Ba2+ ] [ OH– ]2
	[ Ba(OH)2 ]

 

H₃PO_4(aq) <—> 3 H⁺_(aq) + PO₄^3-_(aq)

Ka =	[ H+ ]3 [ PO43- ]
	[ H3PO4 ]

 

However, some acid base equilibria are very confusing. This is because you some times have to imagine the equation with liquid H₂O to make sense. You have to remember that water should always be present in these equations and since H₂O is a liquid it does not count for the equilibrium equation. If this part confuses you, then try going back to the conjugate acid base pairs section in the solutions lesson.

 

Examples: Provide the Ka or Kb equation for the chemical equations below.

 

This is an acid because it produces H⁺ ions

Without water: Li⁺_(aq) <—> LiOH_(aq) + H⁺_(aq)

With water: Li⁺_(aq) + H₂O_(l) <—> LiOH_(aq) + H⁺_(aq)

Ka =	[ H+ ] [ LiOH]
	[ Li+ ]

 

This is a base because it produces OH^– ions

Without water: CH₃COO^–_(aq) <—> CH₃COOH_(aq) + OH^–_(aq)

With water: CH₃COO^–_(aq) + H₂O_(l) <—> CH₃COOH_(aq) + OH^–_(aq)

Kb =	[ CH3COOH ] [ OH– ]
	[ CH3COO– ]

 

You will be expected to predict whether a compound or ion in the reactants acts as an acid or base for future in class questions so you want to get good at it now. The above examples lead us to some clues about the more complex and confusing acid and base equations. First of all, if you have a single positive ion in the reactants, then most likely it is going to act as an acid. Second, if you have a single negative ion in the reactants most likely it is going to act as a base. You can always assume these two things unless you are told otherwise. This allows you to write Ka and Kb equations with minimal information.

 

VIDEO Setting up Ka or Kb Equations Demonstrated Example 1: Give the Ka or Kb formula if nitrate ions are added to water.

 

What are nitrate ions?

Answer: NO₃^–

 

How do you start the chemical equation?

Answer: NO₃^– + H₂O_(l) <—>

 

Is nitrate an acid or base?

Answer: It is a base because it is a negative ion.

 

How do you finish the chemical formula?

Answer: NO₃^–_(aq) + H₂O_(l) <—> OH^–_(aq) + HNO_3(aq)

 

What is the equilibrium equation?

 

COMPLETE ANSWER:

Kb =	[ HNO3 ] [ OH– ]
	[ NO3– ]

 

PRACTICE PROBLEMS: Set up the Ka or Kb equation given the chemical equation below.

 

Mg(OH)_2(aq) <—> Mg²⁺_(aq) + 2 OH^–_(aq)

Answer:

Kb =	[ Mg2+ ] [ OH– ]2
	[ Mg(OH)2 ]

 

K⁺_(aq) + H₂O_(l) <—>

Answer:

Ka =	[ H+ ] [ KOH]
	[ K+ ]

 

S^2-_(aq) <—>

Answer:

Kb =	[ H2S ] [ OH– ]2
	[ S2- ]

 

If carbonate ions are put in water.

Answer:

Kb =	[ OH– ]2 [ H2CO3]
	[ CO32- ]

 

HPO₄^2-_(aq) <—> H₂PO₄^–_(aq) + OH^–_(aq)

Kb =	[ OH– ] [ H2PO4– ]
	[ HPO42-]