Ammonia, NH3 is freely soluble in water, and has the value Kb=1.8x10^-5. If the pH of an ammonia solution is 11.4272, then what original concentration of NH3 (molarity) was used to make up the solution?

NH3 + H2O ==> NH4^+ + OH^-

Kb = (NH4^+)(OH^-)/(NH3)
(x)(x)/(NH3)-x = 1.8 x 10^-5
pH = 11.4272 (where did all these places come from---we only know Kb to 2 places.)
pH = 11.43 pOH = 14-11.43 = 2.57
2.57 = -log(OH^-) and (OH^-) = 2.69 x 10^-3
Substitute into Kb expression and solve for (NH3). You should get something close to 0.4 M but you need to confirm all of the above.

Thank you! I tried all the above and got 0.40 M

To find the original concentration of NH3 (molarity), we can use the relationship between pH and pOH, as well as the expression for the dissociation of NH3 in water.

First, let's find the pOH of the solution using the given pH:

pOH = 14 - pH
pOH = 14 - 11.4272
pOH = 2.5728

Next, we can use the value of pOH to find the concentration of OH- ions in the solution.

pOH = -log[OH-]
2.5728 = -log[OH-]

Taking the antilog of both sides:

[OH-] = 10^-pOH
[OH-] = 10^-2.5728

[OH-] ≈ 0.003187 M

Now, since NH3 is a weak base, it reacts with water to produce OH- ions:

NH3 + H2O ⇌ NH4+ + OH-

The Kb value given (1.8x10^-5) is the equilibrium constant expression for this reaction:

Kb = [NH4+][OH-] / [NH3]

We can assume that the concentration of NH4+ ions is negligible compared to the original concentration of NH3, so we can simplify the equation to:

Kb = [OH-][OH-] / [NH3]
Kb = [OH-]^2 / [NH3]

Substituting the values we found:

1.8x10^-5 = (0.003187)^2 / [NH3]

Now, solving for [NH3]:

[NH3] = (0.003187)^2 / (1.8x10^-5)
[NH3] ≈ 0.056342 M

Therefore, the original concentration of NH3 (molarity) used to make up the solution was approximately 0.056342 M.