What must be the molarity of an aqueous solution of NH3 if it is 4.2% ionized?

Well, let's see, if NH3 is 4.2% ionized, that means it's only partially committed to becoming an NH4+ ion. So, it's a little bit like a commitment-phobe!

But fear not, we can still calculate the molarity. We know that when NH3 ionizes in water, it forms NH4+ and OH- ions. However, since only 4.2% of NH3 is ionized, we can assume the concentration of NH4+ and OH- ions is also 4.2%.

Given this information, we can calculate the molarity by dividing the percentage by 100, and then taking the square root. So, the molarity of the NH3 solution would be approximately the square root of 0.042, or, as I like to call it, the "root of commitment issues."

And remember, the square root of any equation always brings out the fun in math!

To determine the molarity of an aqueous solution of NH3, knowing that it is 4.2% ionized, we need to use the concept of the degree of ionization and equilibrium.

1. Begin by assuming the initial concentration of NH3 in the solution is "C" mol/L.

2. The degree of ionization (α) is defined as the fraction of NH3 that dissociates into NH4+ and OH-. In this case, α = 4.2%.

3. The concentration of NH3 that remains undissociated is (1 - α)C, and the concentration of NH4+ and OH- is (αC) each.

4. In an equilibrium reaction, NH3(aq) ⇄ NH4+(aq) + OH-(aq), the concentration of NH3 is equal to the concentrations of NH4+ and OH-. Therefore, (1 - α)C = αC = NH3 concentration = [NH4+] = [OH-].

5. Since NH3 and NH4+ both contribute to the concentration of NH4+, we can express the concentration of NH4+ in terms of NH3: [NH4+] = αC.

6. Based on the law of chemical equilibrium, the equilibrium constant (Kc) can be expressed as:

Kc = [NH4+][OH-] / [NH3]
= (αC)(αC) / (1 - α)C
= α^2C / (1 - α)

7. In this case, since NH4+ and OH- are generated in a 1:1 ratio, the concentration of OH- is also αC.

8. In a solution of water, the self-ionization constant (Kw) is equal to [H+][OH-], which at 25°C has a constant value of 1.0 x 10^-14.

9. Since [H+] = [OH-], we can substitute the concentration of OH- into the equation for Kw to obtain:

Kw = [H+][OH-]
= (αC)(αC)
= α^2C^2

10. Rearranging the equation above, we get:

α^2C^2 = Kw
C^2 = Kw / α^2
C = √(Kw / α^2)

Now we can substitute the known values:

- Kw = 1.0 x 10^-14 (at 25°C)
- α = 4.2% = 0.042

Calculating C:

C = √(Kw / α^2)
= √(1.0 x 10^-14 / (0.042^2))
= √(1.0 x 10^-14 / 0.001764)
= √(5.667 x 10^-12)
= 7.537 x 10^-6

Therefore, the molarity of the aqueous solution of NH3 is approximately 7.537 x 10^-6 mol/L.

To determine the molarity of an aqueous solution of NH3, we need to use the concept of percent ionization and the dissociation equation of NH3.

NH3 + H2O ⇌ NH4+ + OH-

Given that the solution is 4.2% ionized, it means that only 4.2% of the NH3 molecules have dissociated into NH4+ and OH-. This implies that the concentration of NH4+ in the solution is 4.2% of the initial concentration of NH3.

To calculate the molarity, we need to find the concentration of NH3 in the solution. Let's assume the initial molarity of NH3 is M.

The concentration of NH4+ is then 4.2% of M, which can be expressed as 0.042M. The concentration of OH- is also 4.2% of M, so it is 0.042M.

Since NH3 and NH4+ are in a 1:1 ratio (according to the dissociation equation), the concentration of NH3 is also 0.042M.

Therefore, the molarity of the aqueous solution of NH3 is 0.042M.