So lets consider the most basic condition of 0.10M Ba(OH)2. In solution, 2OH- ions are present. This means that they are twice in concentration to the original concentration since:
Ba(OH)2 -> Ba2 2OH-
Thus, the concentration of OH-, [OH-] is 0.20M.
Calculating pOH = -log[OH-] = -log(0.20M) = 0.70
Calculating pH: 14 = pH pOH; 14 pOH = pH = 14 0.70 = 13.3 (woah! Thats basic!)
To determine the most basic solution, you must find the substances that can give off OH- ions in solution. HF (hydrofluoric acid), HCl (Hydrochloric acid) are NOT bases, do NOT have OH- ions, and thus can be eliminated from the entire options list.
You are left with NaOH, Ba(OH)2 and NH3. All are bases, however NH3 is a weak base (does not fully create OH- ions in solution by reacting with water).
So now you have NaOH and Ba(OH)2. These are both strong bases (dissociate fully in solution to give the metal cation and OH- ions). Now, since they are of same concentration, you must determine which of the two has more OH- ions to give. NaOH has only 1 OH- ion while Ba(OH)2 has 2 OH- ions in solution. Thus, the most basic would be 0.10M Ba(OH)2.
To calculate pH, use the equations shown below. These equations can be derived, but I will not show them here since it requires knowledge of equilibrium and so on:
1) pH = -log[H ] = -log[H3O ]
2) pOH = -log[OH-]
3) 14 = pH pOH
Where [H ] or [H3O ] is the concentration of H ions in solution, [OH-] is the concentration of OH- ions.
So lets consider the most basic condition of 0.10M Ba(OH)2. In solution, 2OH- ions are present. This means that they are twice in concentration to the original concentration since:
Ba(OH)2 -> Ba2 2OH-
Thus, the concentration of OH-, [OH-] is 0.20M.
Calculating pOH = -log[OH-] = -log(0.20M) = 0.70
Calculating pH: 14 = pH pOH; 14 pOH = pH = 14 0.70 = 13.3 (woah! Thats basic!)
Try the other ones. To get [H3O ] or [H ] for HF or [OH-] for NH3 requires you knowing equilibrium which I am not sure if you know of yet (or need to concern yourself with). NH3 generates OH- ions based on the following equilibrium:
NH3 H2O <----> NH4 OH-
This post is last updated on hrtanswers.com at Date : 1st of September – 2022