The measured pH of a 0.100M solution of NH3(aq) at 25C is 11.12. Calculate Kb for Nh3(aq)at 25C.
pH = 11.12. Convert to pOH.
pH + pOH = pKw = 14 so pOH must be 14-11.12 = 2.88
Convert to OH^-
pH = -log(OH^-)
(OH^-) = about 0.0012 but that's just a close estimate. You should redo that estimate for a better answer.
.......NH3 + HOH ==> NH4^+ + OH^-
I......0.1...........0........0
C.......-x...........x........x
E.....0.1-x..........x........x
x = 0.0012 from above. Evaluate the E line, plug that into the Kb expression for Kb and solve for Kb.
Oh ammonia, you never fail to make everyone's nose wrinkle! Now, let's get down to business and calculate Kb for NH3(aq) at 25C.
To begin, we need to determine the pOH of the solution using the measured pH. Since pH + pOH = 14 at 25C (according to the water constant), we can find pOH by subtracting the pH from 14:
pOH = 14 - 11.12
pOH = 2.88
Now, let's convert pOH into OH- concentration by taking the antilog (10 raised to the power of -pOH):
[OH-] = 10^(-2.88)
Next, we recall that NH3(aq) acts as a base in water. It reacts with water to produce OH- ions, forming NH4+ and OH-. The equilibrium expression for this reaction is:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
The equilibrium expression for the base dissociation constant (Kb) is:
Kb = [NH4+][OH-] / [NH3]
Since our solution is 0.100M in NH3, we can substitute the known values into the equation:
Kb = [NH4+][OH-] / 0.100
Now, we have the [OH-] concentration from before. However, we need the concentration of NH4+ to calculate Kb. Since NH4+ is the conjugate acid of NH3, we can assume the concentration is also equal to [OH-]. So our equation becomes:
Kb = ([OH-])^2 / 0.100
Finally, we can plug in the values and calculate Kb:
Kb = (10^(-2.88))^2 / 0.100
And there you have it! Kb for NH3(aq) at 25C. But hey, if you need any more help with chemistry or if you just want to talk about ammonia's nose-pinching capabilities, I'm here for you!
To calculate the base dissociation constant, Kb, for NH3(aq) at 25°C, we can use the equation:
Kw = Ka * Kb
where Kw is the ionization constant of water (1.0 x 10^-14 at 25°C), and Ka is the acid dissociation constant of NH4+.
1. First, let's find the pOH of the solution using the measured pH.
pOH = 14 - pH
pOH = 14 - 11.12
pOH = 2.88
2. Next, let's convert pOH back to OH- concentration using the equation:
pOH = -log[OH-]
[OH-] = 10^(-pOH)
[OH-] = 10^(-2.88)
[OH-] = 1.4 x 10^(-3) M
3. Since NH3 is a weak base, it reacts with water to form OH- ions. In the equilibrium equation:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
The concentration of OH- is given as 1.4 x 10^(-3) M.
The concentration of NH3 and NH4+ are equal to the initial concentration, which is 0.100 M.
4. Now, we can set up an ICE table to determine the equilibrium concentrations:
Initial:
NH3(aq): 0.100 M
NH4+(aq): 0 M
OH-(aq): 1.4 x 10^(-3) M
Change:
NH3(aq): -x
NH4+(aq): +x
OH-(aq): +x
Equilibrium:
NH3(aq): 0.100 - x
NH4+(aq): x
OH-(aq): 1.4 x 10^(-3) + x
5. Since Kb is equal to [NH4+][OH-] / [NH3], we need to determine the equilibrium concentrations.
Kb = ([NH4+][OH-]) / [NH3]
= (x) * (1.4 x 10^(-3) + x) / (0.100 - x)
6. We can assume that the x value (concentration of NH4+ and OH-) is small compared to 0.100 M (concentration of NH3) and can be ignored when subtracted.
Thus, we can simplify the equation to:
Kb = x * (1.4 x 10^(-3)) / 0.100
7. To solve for x, we need to assume that x is small enough that we can neglect it in the calculations.
Since x represents the concentration of NH4+ and OH-, let's assume that x is negligible compared to 0.100 M.
Therefore, we can simplify the equation further:
Kb = (1.4 x 10^(-3)) / 0.100
8. Finally, we can calculate Kb:
Kb = (1.4 x 10^(-3)) / 0.100
Kb ≈ 1.4 x 10^(-2)
Thus, the base dissociation constant, Kb, for NH3(aq) at 25°C is approximately 1.4 x 10^(-2).
To calculate Kb for NH3(aq) at 25°C, we need to use the measured pH of the solution and the concentration of NH3.
Step 1: Convert the pH to pOH.
Since pH + pOH = 14, we can subtract the pH from 14 to get the pOH.
pOH = 14 - 11.12
pOH = 2.88
Step 2: Convert pOH to OH- concentration.
To do this, we need to use the following equation:
pOH = -log[OH-]
To convert pOH to OH-, we can take the antilog of both sides of the equation:
[OH-] = 10^(-pOH)
[OH-] = 10^(-2.88)
Step 3: Determine the concentration of NH3.
The concentration of NH3 is given as 0.100 M.
Step 4: Set up the equation for the Kb expression.
Kb = [OH-][NH3] / [NH4+]
Step 5: Substitute the values into the equation.
Kb = (10^(-2.88))(0.100) / [NH4+]
Step 6: Solve for Kb.
To solve for Kb, we need to know the concentration of NH4+. If it is not provided, we cannot calculate Kb and would need additional information.
Note: The concentration of NH4+ can be calculated if the concentration of NH3 and the pH of the solution are known. However, in this case, the pH of the solution is provided, not the concentration of NH4+. Therefore, it is not possible to calculate Kb with the given information.