Ammonia acts as a base in aqueous solution

[a.a]

Homework Statement

Ammonia acts as a base in aqueous solution; Kb is equal to 1.8 x 10-5. What is the Ka for the ammonium ion NH4+ in aqueous solution?

Homework Equations

The Attempt at a Solution

I am totally lost for this question, can someone please start me off? all i know is that Kw = Ka * Kb

 

[Borek]

all i know is that Kw = Ka * Kb

You are given Kb, you should remember Kw value - just solve for Ka.

 

[Blueshift5]

I can provide some insights into this question. First, let's define some terms. Ammonia (NH3) is a weak base, meaning it does not fully dissociate in water. When ammonia is dissolved in water, it reacts with water molecules to form ammonium ions (NH4+) and hydroxide ions (OH-). This reaction is reversible, meaning that some of the ammonium and hydroxide ions can combine to form ammonia and water again.

The equilibrium constant for this reaction is called the base dissociation constant (Kb). It is a measure of how much ammonia dissociates into ammonium and hydroxide ions in water. In this case, we are given that Kb for ammonia is equal to 1.8 x 10^-5. This means that for every 1 mole of ammonia in solution, 1.8 x 10^-5 moles of ammonium and 1.8 x 10^-5 moles of hydroxide ions are formed.

Now, the question asks for the Ka of the ammonium ion, which is the equilibrium constant for the reaction between ammonium and water to form ammonia and hydronium ions (H3O+). This reaction is also reversible, and the equilibrium constant is related to Kb by the following equation: Kw = Ka * Kb. We know that Kw (the autoionization constant of water) is 1.0 x 10^-14, so we can rearrange the equation to solve for Ka: Ka = Kw / Kb.

Plugging in the value for Kb (1.8 x 10^-5), we get Ka = (1.0 x 10^-14) / (1.8 x 10^-5) = 5.6 x 10^-10. This is the Ka for the ammonium ion in aqueous solution. It tells us how much ammonium dissociates into ammonia and hydronium ions in water.

I hope this explanation helps to clarify the question and guide you in finding the answer. Remember, as a scientist, it's important to understand the concepts behind the equations and not just plug in numbers. Good luck!