the (OH) of ammonia solution is 4.0 x10-2 M. What is the (H3O+) of the solution?

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

initial...0.040...........0........0
change......-x............x.........x
equil.......0.040-x.......x..........x

Kb = (NH4^+)(OH^-)/(NH3)
Substitute Kb, and numbers/symbols from the ICE chart and solve for x = (OH-).
Then (H^+)(OH^-) = Kw = 1E-14.

To find the (H3O+) concentration in the ammonia solution, you need to consider the ionization of water and the reaction between water and ammonia.

First, write the balanced equation for the reaction between water and ammonia:

NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)

Next, write down the expression for the equilibrium constant of this reaction:

Kw = [NH4+][OH-] / [NH3][H2O]

Given that the (OH-) concentration is 4.0 x 10^-2 M, we need to find the (H3O+) concentration. In pure water, the concentration of (OH-) ions is equal to the concentration of (H3O+) ions, and the equilibrium constant (Kw) for water is approximately 1.0 x 10^-14 at 25°C.

So, we can rewrite the expression for Kw in terms of (H3O+) concentration:

Kw = (H3O+)(OH-) = (H3O+)(4.0 x 10^-2)

Since (H3O+) = (OH-) in pure water, we have:

(H3O+) = Kw / (OH-) = (1.0 x 10^-14) / (4.0 x 10^-2)

Calculating this expression, we find that (H3O+) ≈ 2.5 x 10^-13 M.

Therefore, the (H3O+) concentration in the ammonia solution is approximately 2.5 x 10^-13 M.