To find the [H+] concentration in the solution obtained when 0.1 M ammonia is neutralized by a strong HCl, we can use the concept of the ionization constant of water (Kw) and the relationship between Kb and Kw.

The ionization constant of water (Kw) is defined as the product of the concentrations of hydrogen ions ([H+]) and hydroxide ions ([OH-]) in water. At 25°C, Kw is approximately equal to 1.0 x 10^-14.

Ammonia (NH3) is a weak base that reacts with water to form ammonium ions (NH4+) and hydroxide ions (OH-), according to the following equilibrium reaction:

NH3 + H2O ⇌ NH4+ + OH-

The equilibrium constant for this reaction is called the base ionization constant (Kb). The Kb value for ammonia is given as 1.8 x 10^-5.

Since ammonia is a weak base, we can assume that almost all of the NH3 will react with HCl to form NH4+ ions. Therefore, the concentration of NH4+ ions formed will be equal to the initial concentration of ammonia, which is 0.1 M.

Now, let's assume that x moles of NH3 react with x moles of HCl to form x moles of NH4+ ions. Since HCl is a strong acid, it completely ionizes in water to form H+ ions and Cl- ions. Therefore, the concentration of H+ ions formed will also be equal to x moles.

From the equilibrium reaction of ammonia with water, we know that for every mole of NH4+ formed, one mole of OH- is also formed. Therefore, the concentration of OH- ions formed will also be equal to x moles.

Now, using the concept of Kw, we can write the following equation:

Kw = [H+][OH-]

Substituting the values we obtained, we have:

1.0 x 10^-14 = x * x

Simplifying, we have:

1.0 x 10^-14 = x^2

Taking the square root of both sides, we find:

x = 1.0 x 10^-7

Therefore, the concentration of [H+] in the solution obtained when 0.1 M ammonia is neutralized by a strong HCl is 1.0 x 10^-7 M.