Calculate the H+ ion concentration if 0.20M NH3 solution is 2.5% ionized.
0.2 x 0.025 = 0.005
..........NH3 + H2O ==> NH4^+ + OH^-
initial...0.20............0.......0
change....-0.005.......0.005....0.005
equil....0.195.........0.005...0.005
(OH^-) = 0.005.
You know (H^+)(OH^-) = Kw = 1E-14. Substitute and solve for (H^+).
To calculate the H+ ion concentration in an NH3 solution that is 2.5% ionized, we will use the concept of the ionization constant (Ka) of NH3. The equation for the ionization of NH3 is:
NH3 + H2O ⇌ NH4+ + OH-
The ionization constant (Ka) for NH3 is given as:
Ka = [NH4+][OH-] / [NH3]
Given that the NH3 solution is 2.5% ionized, we can assume that x is the concentration of NH3 that ionizes, and 0.20 - x is the concentration of un-ionized NH3.
According to the equation, at equilibrium, the concentration of NH4+ is equal to the concentration of OH-, so:
[NH4+] = [OH-]
Therefore, the equilibrium expression for the given reaction becomes:
Ka = [NH4+][[OH-]] / [NH3] = (x)(x) / (0.20 - x)
Since x represents the concentration of NH3 that ionizes, we can substitute 0.025 (2.5%) for x:
Ka = (0.025)(0.025) / (0.20 - 0.025)
Simplifying:
Ka = 0.000625 / 0.175
Next, we need to solve for the concentration of H+ ions, which is equal to the concentration of NH4+. We'll assume that the concentration of H+ is also equal to x.
So:
[H+] = x = [NH4+] = [OH-]
Finally, we can use the equation for water's ion product constant (Kw) to solve for the OH- concentration:
Kw = [H+][OH-] = 1.0e-14
Given that [H+] = [OH-], we have:
[H+]^2 = 1.0e-14
Taking the square root of both sides:
[H+] ≈ 1.0e-7
Therefore, the concentration of H+ ions in the NH3 solution is approximately 1.0e-7 M.
To calculate the H+ ion concentration in a solution, we need to use the concept of an equilibrium constant, also known as the ionization constant (Ka). In this case, NH3 acts as a weak base that reacts with water to generate OH- ions and its conjugate acid NH4+.
The balanced equation for the reaction is as follows:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Given that the NH3 solution is 2.5% ionized, it means that only 2.5% of NH3 molecules are reacting to produce NH4+ and OH-. Therefore, we can assume that the concentration of NH4+ and OH- ions is 2.5% of the initial concentration of NH3.
First, let's calculate the concentration of NH4+ ions:
NH4+ concentration = (2.5/100) * 0.20 M
NH4+ concentration = 0.005 M
Since the reaction is in equilibrium, the concentration of OH- is equal to the concentration of NH4+. Therefore, OH- concentration is also 0.005 M.
Now, we need to consider that in water, the concentration of H+ ions and OH- ions will always be equal. Therefore, H+ concentration is also 0.005 M.
So, the H+ ion concentration in the 0.20 M NH3 solution, which is 2.5% ionized, is 0.005 M.