compute for the pOH of a solution prepared by dissolving 5.50g KOH in enough to make a 100.0mL solution
...........in enough WHAT? water?
moles KOH = 5.50g/molar mass KOH.
(OH^-) = moles KOH/L soln
pOH = -log(OH^-)
To compute for the pOH of a solution, we first need to determine the concentration of hydroxide ions (OH-) in the solution. We can find this information by using the given mass of KOH and the volume of the solution.
The molar mass of KOH is:
K (potassium) = 39.10 g/mol
O (oxygen) = 16.00 g/mol
H (hydrogen) = 1.01 g/mol
Adding these together will give us:
39.10 + 16.00 + 1.01 = 56.11 g/mol
Next, we need to convert grams of KOH to moles using the molar mass:
5.50 g KOH * (1 mol KOH / 56.11 g KOH) = 0.09805 mol KOH
Since the volume of the solution is given in milliliters, we need to convert it to liters:
100.0 mL * (1 L / 1000 mL) = 0.100 L
Now we can find the concentration of KOH in the solution (in units of mol/L or M):
0.09805 mol / 0.100 L = 0.9805 M KOH
Since KOH is a strong base, it will completely dissociate in water, meaning the concentration of hydroxide ions (OH-) is the same as the concentration of KOH.
Finally, we can calculate the pOH using the concentration of OH-:
pOH = -log[OH-]
pOH = -log[0.9805]
Using a scientific calculator, we find that pOH ≈ 0.007
Therefore, the pOH of the solution prepared by dissolving 5.50g KOH in enough to make a 100.0mL solution is approximately 0.007.