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)
It's 11.5
i got 11.5 but im not sure because the problems i did before i had to change to ka first and that was the pH and got those answers right...but i don't understand why i have to do it this way though its the right way...quest is saying its wrong..
To determine the pH of the ammonia solution, we need to calculate the concentration of hydroxide ions (OH-) in the solution, which can be done using the Kb (base ionization constant) of ammonia.
The Kb expression for ammonia can be written as follows:
Kb = [NH4+][OH-] / [NH3]
We know the concentration of ammonia (NH3) is 0.563 M. However, at room temperature, ammonia predominantly exists as NH3 rather than as NH4+ and OH-. Therefore, we can assume the concentration of NH4+ is negligible compared to the concentration of NH3.
Now, using the Kb expression and the molar concentration of ammonia, we can solve for OH- concentration.
Kb = [NH4+][OH-] / [NH3]
1.78 × 10^-5 = [OH-] * 0.563 / 0.563
1.78 × 10^-5 = [OH-]
Therefore, the concentration of OH- ions in the solution is 1.78 × 10^-5 M.
To find the pH, we need to convert the concentration of OH- ions to the concentration of H+ ions using the equation Kw = [H+][OH-].
At room temperature, the value of Kw (ion product of water) is 1.0 × 10^-14.
1.0 × 10^-14 = [H+][1.78 × 10^-5]
[H+] = 1.0 × 10^-14 / 1.78 × 10^-5
[H+] = 5.62 × 10^-10 M
To find the pH, we use the equation pH = -log[H+].
pH = -log(5.62 × 10^-10)
pH = 9.25
Therefore, the pH of the 0.563 M ammonia solution would be approximately 9.25.
Based on the given answer choices, the closest option would be 9, not 4.5.
NH3 + H2O ==> NH4^+ + OH^-
0.563..........0........0
-x..............x........x
0.563-x.........x........x
Kb = 1.68E-5 = (x)(x)/(.563-x)
Solve for x, which is OH^-, convert to pOH, then to pH. I agree that the answer isn't 4.5