What is the equilibrium concentration of ammonium ion in a 0.77 M solution of ammonia (NH3, Kb = 1.8 ´ 10–5) at 25oC?

NH3 + HOH ==> NH4^+ + OH^-

Set up an ICE chart, plug into the Kb expression for NH3 and solve. Post your work if you get stuck.

To calculate the equilibrium concentration of ammonium ion (NH4+) in a solution of ammonia (NH3), we need to use the concept of base dissociation equilibrium and the equilibrium constant (Kb).

The base dissociation equation for ammonia is as follows:
NH3 + H2O ⇌ NH4+ + OH-

The equilibrium constant expression, Kb, for this reaction at 25°C is given as:
Kb = [NH4+][OH-] / [NH3]

Given:
- Concentration of ammonia (NH3) = 0.77 M
- Kb = 1.8 × 10^(-5)

Let's assume the concentration of NH4+ at equilibrium is represented as x M.

At equilibrium:
- [NH3] = initial concentration - change in concentration = 0.77 - x (M)
- [NH4+] = x (M)
- [OH-] = x (M)

Since the concentration of OH- is also x, we can substitute these values into the Kb expression:

Kb = [NH4+][OH-] / [NH3]
1.8 × 10^(-5) = (x)(x) / (0.77 - x)

Now, solve for x by rearranging the equation and solving the quadratic equation:

1.8 × 10^(-5) = x^2 / (0.77 - x)

Rearrange the equation to obtain:
x^2 = 1.8 × 10^(-5) (0.77 - x)

Simplify:
x^2 = 1.386 × 10^(-5) - 1.8 × 10^(-5)x

Rearrange again:
x^2 + 1.8 × 10^(-5)x - 1.386 × 10^(-5) = 0

Now, solve the quadratic equation using the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / 2a

Substituting the values:
a = 1
b = 1.8 × 10^(-5)
c = -1.386 × 10^(-5)

x = [-1.8 × 10^(-5) ± √((1.8 × 10^(-5))^2 - 4(1)(-1.386 × 10^(-5)))] / 2(1)

After calculating, you will have two possible values for x. Consider only the positive value as [NH4+] concentration cannot be negative.

The positive value of x will represent the equilibrium concentration of NH4+ in the solution of ammonia at 25°C.