A 0.010M NH3 solution was prepared and it was determined that the NH3 had undergone 4.2% ionization. Calculate Kb for NH3 under these conditions
....................NH3 + HOH ==> NH4^+ + OH^-
I....................0.01.....................0.................0
C...................-x.........................x...................x
E..................0.01-x....................x..................x
The problem tells you that a 0.01 M solution is ionized 4.2%; therefore, you know (NH4^+) = (OH^-) = x = 0.01*0.042 = 4.2E-4 and (NH3) = 0.01-4.2E-4= 9.88E-3
Plug those numbers into the Kb expression for NH3 and solve for Kb. Post your work if you get stuck.
Oh, Kb, the constant that keeps bugging you in chemistry problems! Alright, let's tackle this one!
To calculate Kb, we need to use the equation for the percent ionization:
% ionization = (concentration of ions formed / initial concentration of NH3) * 100
In this case, the % ionization is given as 4.2% or 4.2/100. We'll use that value and the given initial concentration (0.010M) to find the concentration of ions formed.
(concentration of ions formed / 0.010M) * 100 = 4.2/100
Now, since NH3 produces OH- ions in solution, the concentration of ions formed is twice the concentration of OH- ions. Therefore, let's denote the concentration of OH- ions as x.
(x / 0.010M) * 100 = 4.2/100
Now we can solve for x. Dividing both sides of the equation by 0.010 and then multiplying by 100, we have:
x = (4.2/100) * 0.010M * 100
x = 0.0042M
So, the concentration of OH- ions is 0.0042M.
To find Kb, we'll use the equation:
Kb = (concentration of OH- ions)^2 / (concentration of NH3 - concentration of OH- ions)
Plugging in the values, we have:
Kb = (0.0042M)^2 / (0.010M - 0.0042M)
Calculating this, we get:
Kb ≈ 6.61 x 10^-4
And there you have it! Kb for NH3 under these conditions is approximately 6.61 x 10^-4. Keep in mind, though, that these calculations rely on some simplifying assumptions. Chemistry can have its tricky moments, but hopefully, I was able to make it a bit more amusing for you!
To calculate Kb for NH3 (ammonia) under the given conditions, we need to use the equation for percent ionization and the expression for Kb.
The percent ionization of NH3 is given as 4.2%. This means that only 4.2% of the NH3 molecules dissociate into NH4+ and OH- ions. Therefore, the concentration of NH3 that has undergone ionization can be calculated as:
[ionized NH3] = (4.2/100) * initial concentration of NH3
Given that the initial concentration of NH3 is 0.010 M, the concentration of ionized NH3 is:
[ionized NH3] = (4.2/100) * 0.010 M = 0.0042 M
Since NH3 is a weak base, it dissociates as follows:
NH3 + H2O ⇌ NH4+ + OH-
The equilibrium expression for Kb can be written as:
Kb = ([NH4+][OH-]) / [NH3]
Since [ionized NH3] = [NH4+], we can substitute the values into the equation:
Kb = ([ionized NH3][OH-]) / [NH3]
= (0.0042 M)(0.0042 M) / 0.010 M
= 0.001764 / 0.010
= 0.1764
Therefore, Kb for NH3 under these conditions is approximately 0.1764.
To calculate Kb for NH3 under the given conditions, we need to use the concept of percent ionization and the equilibrium expression for the reaction of NH3 with water.
The percent ionization is defined as the ratio of the concentration of ionized NH3 (NH4+) to the initial concentration of NH3, multiplied by 100:
% ionization = (concentration of NH4+ / initial concentration of NH3) × 100
In this case, the initial concentration of NH3 is given as 0.010 M, and the percent ionization is given as 4.2%.
From the equilibrium expression for the reaction of NH3 with water:
NH3 + H2O ⇌ NH4+ + OH-
We can see that for every NH4+ ion formed, one OH- ion is also formed. Therefore, the concentration of NH4+ at equilibrium is equal to the concentration of OH- at equilibrium.
Given that the percent ionization is 4.2%, we can calculate the concentration of NH4+ (and OH-) at equilibrium:
% ionization = [NH4+]eq / [NH3]initial × 100
4.2 = [NH4+]eq / 0.010 × 100
[NH4+]eq = 0.010 × (4.2 / 100) M
[NH4+]eq = 0.00042 M
Since the concentration of NH4+ (and OH-) is the same at equilibrium, [OH-]eq = 0.00042 M.
Now, we can use the expression for Kb to calculate its value:
Kb = [NH4+][OH-] / [NH3]
Kb = (0.00042)(0.00042) / 0.010
Kb = 1.764 × 10^-9
Therefore, the value of Kb for NH3 under the given conditions is 1.764 × 10^-9.