What would be the pH of a 0.563 M solution of ammonia (NH3) at room temperature? The Kb of ammonia is 1.78 × 10−5.
1. 2.5
2. 7
3. 9
4. 4.5
5. 11.5
( I did this problem but apparently 4.5 isnt the answer)
See your other post above.
To find the pH of a solution of ammonia, we need to calculate the concentration of hydroxide ions (OH-) and then convert it to the pH scale.
Ammonia (NH3) is a weak base, and it reacts with water to produce hydroxide ions (OH-) and ammonium ions (NH4+):
NH3 + H2O ⇌ NH4+ + OH-
The equilibrium constant for this reaction is the base dissociation constant (Kb), which is given as 1.78 × 10^-5.
To find the concentration of OH- ions, we can set up an ICE table:
NH3 + H2O ⇌ NH4+ + OH-
Initial: 0.563 0 0
Change: -x -x +x
Equilibrium: 0.563 - x -x x
The equilibrium expression for Kb is:
Kb = [NH4+][OH-] / [NH3]
Kb = (x)(x) / (0.563 - x)
Since the initial concentration of NH4+ and OH- is negligible, we can approximate them as zero. This simplifies the equation to:
Kb ≈ x^2 / 0.563
Now we can solve for x, which represents the concentration of OH- ions. Rearranging the equation, we have:
x^2 = Kb * (0.563)
x = sqrt(Kb * (0.563))
Plugging in the values, we get:
x = sqrt[(1.78 × 10^-5) * (0.563)]
Calculating the square root of this value will give us the concentration of OH- ions in the solution. To convert this to pH, we need to use the equation:
pOH = -log10[OH-]
pH = 14 - pOH
Calculating pOH and pH will provide the answer to the question.