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.