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?

KaKb=Kw

To find the Ka for the ammonium ion (NH4+) in aqueous solution, we can use the relationship between Ka and Kb for the conjugate acid-base pair.

The Kb for ammonia (NH3) is given as 1.8 x 10^-5. Since ammonia is a weak base, its conjugate acid, NH4+, will be a weak acid.

The relationship between Ka and Kb for a conjugate acid-base pair is given by:

Ka x Kb = Kw

Where Kw is the ionization constant for water, which is equal to 1.0 x 10^-14 at 25°C.

Rearranging the equation, we get:

Ka = Kw / Kb

Substituting the values, we have:

Ka = (1.0 x 10^-14) / (1.8 x 10^-5)

Calculating this, we find:

Ka ≈ 5.56 x 10^-10

Therefore, the Ka for the ammonium ion (NH4+) in aqueous solution is approximately equal to 5.56 x 10^-10.

To determine the Ka for the ammonium ion NH4+ in aqueous solution, we need to use the relationship between Ka and Kb for a conjugate acid-base pair.

Since ammonia (NH3) acts as a base, its conjugate acid is the ammonium ion (NH4+). The Kb value given for ammonia is 1.8 x 10^-5.

The relationship between Ka and Kb is given by the equation:

Kw = Ka × Kb

Where Kw is the ion product of water, which is equal to 1.0 x 10^-14 at 25°C.

To find Ka, we need to rearrange the equation:

Ka = Kw / Kb

Now, we can substitute the known value of Kb into the equation:

Ka = (1.0 x 10^-14) / (1.8 x 10^-5)

Calculating this equation gives the value of Ka for the ammonium ion NH4+ in aqueous solution.