Question
Determine the pH of 0.15M ammonia with a kb=1.8×10 ^-5
Answers
Answered by
GPT 3.5
To find the pH of the 0.15M ammonia solution, we first need to calculate the concentration of hydroxide ions (OH-) produced when ammonia reacts with water.
The equation for the reaction of ammonia (NH3) with water (H2O) is:
NH3 + H2O ⇌ NH4+ + OH-
From this, we can see that ammonia acts as a base and accepts a proton (H+) from water to form the ammonium ion (NH4+) and hydroxide ion (OH-).
To find the concentration of OH- produced, we need to first calculate the concentration of NH4+ formed. Since ammonia dissociates in water to form ammonium and hydroxide ions in a 1:1 ratio, the concentration of NH4+ will be the same as the concentration of OH- produced.
Let x be the concentration of NH4+ (equivalent to OH-), then the initial concentration of NH3 is 0.15M and the initial concentration of OH- is 0M.
Using the equilibrium constant expression for the reaction:
Kb = [NH4+][OH-] / [NH3]
1.8x10^-5 = x * x / 0.15
x^2 = 1.8x10^-5 * 0.15
x^2 = 2.7x10^-6
x = sqrt(2.7x10^-6)
x ≈ 0.00164
So, the concentration of OH- (and also NH4+) is approximately 0.00164M.
Now, to find the pOH of the solution:
pOH = -log[OH-]
pOH = -log(0.00164)
pOH ≈ 2.79
Finally, to find the pH of the solution:
pH = 14 - pOH
pH = 14 - 2.79
pH ≈ 11.21
Therefore, the pH of the 0.15M ammonia solution is approximately 11.21.
The equation for the reaction of ammonia (NH3) with water (H2O) is:
NH3 + H2O ⇌ NH4+ + OH-
From this, we can see that ammonia acts as a base and accepts a proton (H+) from water to form the ammonium ion (NH4+) and hydroxide ion (OH-).
To find the concentration of OH- produced, we need to first calculate the concentration of NH4+ formed. Since ammonia dissociates in water to form ammonium and hydroxide ions in a 1:1 ratio, the concentration of NH4+ will be the same as the concentration of OH- produced.
Let x be the concentration of NH4+ (equivalent to OH-), then the initial concentration of NH3 is 0.15M and the initial concentration of OH- is 0M.
Using the equilibrium constant expression for the reaction:
Kb = [NH4+][OH-] / [NH3]
1.8x10^-5 = x * x / 0.15
x^2 = 1.8x10^-5 * 0.15
x^2 = 2.7x10^-6
x = sqrt(2.7x10^-6)
x ≈ 0.00164
So, the concentration of OH- (and also NH4+) is approximately 0.00164M.
Now, to find the pOH of the solution:
pOH = -log[OH-]
pOH = -log(0.00164)
pOH ≈ 2.79
Finally, to find the pH of the solution:
pH = 14 - pOH
pH = 14 - 2.79
pH ≈ 11.21
Therefore, the pH of the 0.15M ammonia solution is approximately 11.21.